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ANALYTICAL MECHANICS
FOR ENGINEERS
BY
FRED B. SEELY, M.S.
Professor of Theoretical and Applied Mechanics
AND
NEWTON E. ENSIGN, A.B., B.S.
Assistant Professor of Theoretical and Applied Mechanics
University of Illinois
TOTAL ISSUE, TEN THOUSAND
NEW YORK
JOHN WILEY & SONS, Inc
London: CHAPMAN & HALL, Limited
1921
UNIVERSITY OF CALIFORNIA
LIBRARY •
BRANCH OF THE
COLLEGE OF AGRICULTURE
Copyright, 1921,
By FRED B. SEELY
and
NEWTON E. ENSIGN
PRESS OP
BRAUNWORTH ft CO>
tf /jl BOOK MANUFACTURERS
BROOKLYN, N. Y.
PREFACE
This book, as its name suggests, presents those principles
of mechanics that are believed to be essential for the student of
engineering.
Throughout the book the aim has been to make the principles
of mechanics stand out clearly ; to build them up as much as pos-
sible from common experience (the student's experience); to
apply the principles to concrete problems of practical value;
and to emphasize the physical rather than the mathematical
interpretation of the principles. Important equations are printed
in bold-faced type and the statements of the more important
principles are italicized.
The book is divided into three parts; namely, Statics, Kine-
matics, and Kinetics. Statics is presented first because of its
simpUcity and its direct relation to the student's experience.
However, in the first two chapters are developed certain concepts
and elementary principles that are fully as important in kinetics
as in statics, and the authors feel that it is essential to a satis-
factory grasp of mechanics, as a whole, that sufficient time and
care be taken to cause these elementary concepts and principles
to crystallize in the student's mind before the more general prin-
ciples and problems are studied. The equilibrium of the various
types of force systems are treated both by the algebraic and by the
graphical method. A large number of problems involving the
equilibrium of the simpler structures and machines are given, and
figures illustrating the structures and machines are used freely.
Although kinematics as herein developed is mainly a prehm-
inary to kinetics, the authors' experience indicates that the
kinematic properties of motion must be isolated and developed
with care if they are to be used with success in the study of the
kinetics of the motion.
Z3Zf<^
iv PREFACE
Both kinematics and kinetics have been developed with regard
for the increasing importance of dynamics to engineers. The
geometric and physical conceptions and interpretations of the
quantities in Icinematics have been emphasized rather than the
mathematical conceptions. A treatment of acceleration is given
which, it is hoped, will help to overcome some of the difficulties
frequently found in the use of this quantity. The treatment of
kinetics has been restricted to the more common types of motion
found in engineering practice, but these motions have been treated
more fully than is usual in elementary texts on mechanics. This
is particularly true of plane motion. D'Alembert's principle
(involving inertia forces) has been used for each type of motion
as a second method of solution. The methods of procedure used
in the analysis of kinetics problems are strongly emphasized both
in the general discussions and in the solutions of illustrative
problems.
Illustrative problems are given at the end of the more important
articles and many problems are offered for solution. Great care
has been exercised in selecting problems that are of practical
interest and yet are easily comprehended and are free from unim-
portant details so that the principles used in their solution will
stand out clearly. The answers to about one-half of the problems
are given.
Graphical methods of representation and of solution have been
used frequently in all three parts of the book. A knowledge of
elementary calculus is assumed although little use is made of it
in the first four chapters.
The discussion of centroids (Chapter V) is developed directly
from the principle of moments — a principle given much emphasis
throughout the book.
Several special topics are discussed in Section 3 of Chapter IX.
They may be omitted without interfering with the continuity of
the book, or any one of the topics may be studied alone without
studying the whole section. Further, if it is desired to reduce the
time given to kinetics, the second method of analysis of the motion
of rotation or of plane motion (which employs D'Alembert's
principle and inertia forces) in Section 2 of Chapter IX may be
omitted. And, in general, the last part of the material in any
section or chapter may be omitted without interfering with the
student's progress in the first part of the next section or chapter.
PREFACE V
The authors wish to acknowledge their indebtedness to Pro-
fessor A. N. Talbot for his interest in the book during its prepara-
tion and for his helpful suggestions on the treatment of certain
important topics. The authors are also indebted to Professor
G. A. Goodenough for valuable material in the treatment of Gov-
ernors and to Professor 0. A. Leutwiler for the data and figures
for a number of valuable engineering problems.
F. B. S.
N. E. E.
Urbana, Illinois,
December, 1920.
CONTENTS
PAGE
Preface iii
PART I. STATICS
Chapter I. Fundamental Conceptions and Definitions
1. Introduction 1
2. RigidBody 3
3. Conception of a Force 3
4. External Effects of a Force 4
5. Characteristics of a Force 4
6. Measure of a Force. Units 4
7. Scalar and Vector Quantities. Vector Representation of a Force. . . 6
8. Classification of Forces. Definitions 7
9. Principle of Transmissibility 8
10. Parallelogram Law 8
11. Triangle Law 9
12. Resolution of a Force 11
13. Moment of a Force 14
14. Principle of Moments. Varignon's Theorem 16
15. Couples 19
16. Characteristics of a Couple 20
17. Vector Representation of a Couple 20
18. Resolution of a Force into a Force and a Couple 22
19. Methods of Solution of Problems 24
Chapter II. Resultants op Force Systems
20. Introduction 28
§ 1. Collinear Forces
21. Algebraic Method 28
§ 2. Concurrent Forces in a Plane
22. Graphical Methods 29
23. Algebraic Method 31
vii
viil CONTENTS
§ 3, Parallel Forces in a Plane
PAGB
24. Graphical Methods 34
25. Principle of Moments 38
26. Algebraic Method 40
27. Couples 44
§ 4. Non-concurrent, Non-parallel Forces in a Plane
28. Graphical Methods 48
29. Principle of Moments 49
30. Algebraic Method 50
§ 5. Concurrent Forces in Space
31. Graphical Method 53
32. Algebraic Method 54
§ 6. Parallel Forces in Space
33. Graphical Method 56
34. Principle of Moments 57
35. Algebraic Method 57
§ 7. Couples in Space
36. Resultant of a System of Couples 60
37. Composition of Couples by Means of Vectors 62
38. Resolution of Couples 64
§ 8. Non-concurrent, Non-parallel Forces in Space
39. Graphical Method 65
40. Principle of Moments 65
41. Algebraic Method 66
Chapter III. Equilibrium of Force Systems
§ 1. Introduction
42. Preliminary 69
43. Graphical Conditions of Equilibrium 70
44. Algebraic Conditions of Equilibrium 70
§ 2. Collinear Forces
45. Equations of EquiUbrium 72
§ 3. Concurrent Forces in a Plane
46. Equations of Equilibrium 73
47 Lami's Theorem 74
48 Three Forces in Equilibrium 75
§ 4. Parallel Forces in a Plane
49. Equations of Equilibrium 78
CONTENTS ix
§5. N cm-concurrent, Non-parallel Forces in a Plane —
PAGE
50. Equations of Equilibrium 82
51. Graphical Solution of a Typical Problem 85
52. Choice of Moment-centers and of Directions of Resolution 87
§ 6. Equilibrium of Trusses, Cranes, and Cables
53. Stresses in Trusses 88
54. Method of Joints 89
55. Method of Sections 90
56. Graphical Analysis of Trusses • 95
57. Cranes 98
58. Stresses in Flexible Cables 102
59. The Parabohc Cable 102
60. The Catenary 106
§ 7. Concurrent Forces in Space
61. Equations of Equilibrium 110
§ 8. Parallel Forces in Space
62. Equations of Equilibrium 112
§ 9. Non-concurrent, Non-parallel Forces in Space
63. Equations of Equilibrium 114
Chapter IV. Friction
64. Friction Defined 127
65. Coefficient of Friction 128
66. Angle of Friction and the Friction Cone 129
67. Angle of Repose 133
68. The Laws of Friction 134
69. The Inclined Plane 135
70. The Wedge 138
71. The Screw 142
72. Journal Friction and the Friction Circle 145
73. Pivot Friction . 148
74. SoUd Flat Pivot 149
75. Hollow Flat Pivot or Collar Bearing 150
76. Conical Pivot 151
77. Rolling Resistance 153
78 Belt Friction 155
Chapter V. First Moments and Centroids
79. First Moments 160
80. Centroids 160
81. Planes and Lines of Symmetry 162
82. Centroids by Integration. . 163
X CONTENTS
83. Centroids of Composite Figures and Bodies 168
84. Theorems of Pappus and Guldinus 171
85. Center of Gravity and Mass-center 173.
86. Graphical Method of Determining Centroids of Areas 177
87. Determination of Center of Gravity by Experiment 178
Chapter VI. Second Moment, Moment of Inertia
§ 1. Moments of Inertia of Areas
88. Moment of Inertia of an Area Defined 180
89. Polar Moment of Inertia 181
90. Radius of Gyration 182
91. Parallel Axis Theorem for Areas 183
92. Moments of Inertia by Integration 184
93. Moments of Inertia of Composite Areas 190
94. Product of Inertia Defined 194
95. Axes of Symmetry 194
96. Parallel Axis Theorem for Products of Inertia 196
97. Moments of Inertia with Respect to Inclined Axes 198
98. Principal Axes 199
99. Moments of Inertia of Areas by Graphical and Approximate Methods 202
§ 2. Moments of Inertia of Bodies
100. Moment of Inertia of Mass Defined 205
101. Radius of Gyration 207
102. Parallel Axis Theorem for Masses 207
103. Moments of Inertia with Respect to Two Perpendicular Planes 208
104. Moments of Inertia of Simple Solids by Integration 209
105. Moments of Inertia of Composite Bodies 2J18
106. Moments of Inertia of Bodies by Experimental Methods 220
PART 11. KINEMATICS
107. Introduction 221
Chapter VII. Motion of a Particle
108. Types of Motion 222
109. Linear Displacement 222
110. Angular Displacement 224
111. Relation between Linear and Angular Displacements 225
112. Linear Velocity and Speed 226
113. Distance-time Graph 228
114. Angular Velocity 229
115. Relation between Linear and Angular Velocities 230
116. Components of Velocity 231
CONTENTS xi
rAOB
117. Linear Acceleration 7~; .^rr 335
118. Tangential and Normal Accelerations 236
119. Speed-time Graph 241
120. Angular Acceleration 242
121. Relation between Linear and Angular Accelerations; 243
122. Axial Components of Acceleration 245
123. Motion of a Projectile 248
124. Uniformly Accelerated Rectilinear Motion 250
125. Uniformly Accelerated Circular Motion 251
>• 126. Simple Harmonic Motion 253
ll27. Non-uniformly Accelerated Motion 256
128. Relative Motion 262
129. Transverse and Radial Components of Acceleration 269
130. Coriolis' Law 270
Chapter VIII. Motion of Rigid Bodies
131. Introduction 276
132. Translation 276
133. Rotation 277
134. Plane Motion 278
135. Instantaneous Center. 284
PART III. KINETICS
Chapter IX. — Force, Mass, and Acceleration
§ 1. Preliminary Considerations. Kinetics of a Particle
136. Introduction 287
137. The General Kinetics Problem 288
138. Characteristics of a Force System 288
139. Inertia and Mass 289
140. Newton's Laws 293
141. Mathematical Statement of Newton's Second Law. Units 294
142. Other Systems of Units 296
143. Equations of Motion for a Particle 297
§ 2. Kinetics of Bodies
144. Tntroduction. Methods of Analysis 303
Translation
145. Kinetics of a Translating Rigid Body 305
Rotation
146. Kinetics of a Rotating Rigid Body : 313
147. Second Method of Analysis. Inertia Forces 325
148. Center of Percussion 336
XU CONTENTS
Plane Motion
^ PAGE
149. Kinetics of Plane Motion of a Rigid Body 336
150. Second Method of Analysis. Inertia Forces 345
151. Limitations of the Moment Equation, T = Ia 351
152. Principle of the Motion of the Mass-center 353
§ 3. Special Topics in Kinetics
153. Introduction 356
154. Hoop Tension in Flywheels 356
155. Superelevation of Railway Track 358
Pendulums
156. Simple Circular Pendulum 360
157. Compound Pendulum 363
158. Center of Oscillation 364
159. Torsion Pendulum 365
160. Experimental Determination of Moment of Inertia 367
Balancing
161. Need for Balancing 369
162. Balancing of Rotating Masses 370
163. Several Masses in a Single Plane of Rotation 372
164. Masses in Different Transverse Planes 373
The Gyroscope
165. The Problem Defined 378
166. Analysis of Forces in the Gyroscope 379
167. The Moment of the Gyroscopic Couple 382
Governors
168. The Action of Governors 386
169. The Conical Pendulum 387
170. The Loaded Governor 388
171. The Porter Governor 389
172. The Centrifugal Shaft Governor 391
173. The Inertia Shaft Governor 392
174. Comparison of the Two Types of Governors 393
175. Analysis of Forces in the Rite's Inertia Governor 393
Chapter X. — Work and Energy
176. Introduction 405
§ 1. Work and Power
177. Work Defined 406
178. Algebraic Expressions for Work Done by a Force 406
179. Work Done by a Couple , 408
180. Work a Scalar Quantity. Sign and Units of Work 409
181. Graphical Representation and Calculation of Work 409
CONTENTS xiii
_^ PAGE
182. Work Done on a Body by a Force System 414
183. Power Defined 417
184. Units of Power 417
185. Special Equations for Power 418
§2. Energy
186. Energy Defined 420
187. Potential Energy 421
188. Kinetic Energy 422
189. Kinetic Energy of a Particle 422
190. Kinetic Energy of a Body 424
191. Non-mechanical Energy 428
§ 3. Principles of Work and Energy
192. Preliminary 430
193. Principle of Work and Kinetic Energy 430
194. Application of Principle to Special Cases of Motion of Rigid Bodies. 432
195. Conservation of Energy 439
196. Relation between Potential and Kinetic Energy for Conservative
Systems 440
§ 4. Efficiency. Dissipation of Energy
197. Efficiency Defined 442
198. Dissipation of Energy 443
199. Work Lost Due to Friction in Bearings 444
200. A Simple Dynamometer. Prony Brake 446
Chapter XI. — Impulse and Momentum
201. Preliminary 449
§ 1. Impulse
202. Impulse and Impact Defined. Units 450
203. Components of Linear Impulse 451
204. Moment of Impulse. Angular Impulse 452
§ 2. Momentum
205. Momentum of a Particle Defined. Units 453
206. Components of Momentum. Moment of Momentum 454
207. Linear Momentum of a Body 455
208. Angular Momentum of a Rotating Rigid Body 456
209. Angular Momentum of a Rigid Body Having Plane Motion 457
§ 3. Principles of Impulse and Momentum
210. Preliminary 459
211. Principle of Linear Impulse and Linear Momentum 459
212. Principle of Angular Impulse and Angular Momentum 461
xiv CONTENTS
PAGE
213. Method of Analysis of the Motion of a Body by Means of Impulse
and Momentum 462
214. Application of the Principles of Impulse and Momentum to Special
Types of Motion of Rigid Bodies 463
215. Conservation of Momentum 467
216. Impact 471
217. Impact of Two Translating Bodies . 473
218. Impact of a Translating Body with a Rotating Body 475
219. Loss of Kinetic Energy in Direct Central Impact 476
220. Angular Momentum a Vector Quantity 477
221. The Gyroscopic Couple 478
ANALYTICAL MECHANICS FOR ENGINEERS
PART I. STATICS
CHAPTER I
FUNDAMENTAL CONCEPTIONS AND DEFINITIONS
1. Introduction. — ^The term Mechanics is used in a broad
sense to denote the science which treats of the motion of bodies,
rest being considered as a special case of motion. The science
of mechanics constitutes a large part of our knowledge of the
laws of the Universe, including the laws concerning gases and
liquids as weU as those of rigid bodies, and it takes a prominent
place in the study of Astronomy and Phj^sics as well as in the
study of machines and structures which are involved in engineer-
ing practice.
The object of Analytical Mechanics as developed in this book
is to determine the laws by which the motions (including the state
of rest) of bodies (mainly rigid bodies) are governed and to apply
these laws to conditions met in engineering practice.
In the development of the laws of mechanics certain concepts
are assumed to be fundamental, that is, no one of them can be
expressed in terms of the others or in simpler terms. Such
concepts grow out of our experiences, and other ideas and laws
are derived from these condensed experiences.
The fundamental concepts involved in the laws of mechanics
of rigid bodies are: (1) force, which is made known to us through
the tension and the compression of our muscles as a pull or a push,
(2) bodies or inert material (matter) on which forces act and with-
out which forces cannot exist, (3) space, and (4) time. A more
definite understanding of force and inert bodies can be obtained
after the laws of kinetics have been developed. To start with,
2 FUNDAMENTAL CONCEPTIONS AND DEFINITIONS
however, it is necessary only to recognize the existence of these
quantities.
In the process of the development of the laws of mechanics,
considerable use is made of mathematics. It should be kept in
mind, however, that mechanics is a physical science and that
mathematics is used, mainly, as a tool to express and interpret
physical laws.
For convenience, the study of mechanics is considered under
three main divisions; namely. Statics, Kinematics , and Kinetics,
Statics is that branch of mechanics which treats of bodies that are
acted on by balanced forces and hence are at rest or are moving
with uniform motion (without change of motion). Kinematics
is that branch of mechanics which treats of the motion of bodies
without considering the manner in which the influencing factors
(force and matter) affect the motion. It deals with the funda-
mental concepts of space and time, and the quantities, velocity
and acceleration, derived therefrom. It is, therefore, sometimes
called the geometry of motion. Kinematics forms an important
part of the study of mechanics, not only because it treats of a part
of the general kinetics problem, but also because in many problems
which involve mainly the relative motion of parts of a machine,
the principles of kinematics, alone, are sufficient for the solution
of the main problem. Such problems are discussed in treatises on
Kinematics of Machinery, in which subject the motion of such
machine elements as valve gears, quick-return mechanisms, etc.,
are considered. Kinetics is that branch of mechanics which treats
of bodies which are acted on by unbalanced forces and, hence,
have non-uniform or accelerated motions. In particular, it treats
of the change of motion of bodies and the manner in which the
change is related to the factors that affect it; namely, the actions
of other bodies (forces), and the properties (inertia, etc.) of the
bodies themselves. It will be noted that the study of the motion
of a body that moves uniformly may be regarded either as a prob-
lem in statics or as ^ special case of a kinetics problem.
Since both statics and kinetics deal with the action of forces on
bodies, that part of mechanics embraced in these two subdivisions
of the subject is sometimes called Dynamics. Frequently, how-
ever, the term dynamics is used in technical literature to denote
those subdivisions of mechanics with which the idea of motion is
most closely associated, namely, kinematics and kinetics.
CONCEPTION OF A FORCE 3
For simplicity the kinematics and kinetics of a particle (mate-,
rial point) will be treated before extending the study to bodies.
In many problems the body considered may be assumed, without
serious error, to be a particle. This assumption may always be
made when the dimensions of the body are negligible in comparison
with the range of its motion. A body, however, may always be
considered to be made up of particles.
2. Rigid Body. — As was stated in Art. 1, the bodies dealt with
in this book are, in the main, considered to be rigid. A rigid body
is defined as a definite portion of matter the parts (particles) of
which do not move relative to each other. Actual soHd bodies
are never rigid. The relative motion (deformation) of their par-
ticles forms an important part of the study of Strength of Materials.
But the theoretical laws which govern the motion of ideal rigid
bodies may be used, usually with very small error or, with modi-
fications if necessary, to determine the motion of actual solid
bodies.
3. Conception of a Force. — It was stated in Art. 1 that force
is one of the fundamental concepts on which the subject of me-
chanics of rigid bodies is built. A force is the action of one body on
another body. The idea of force, then, implies the mutual actions
of two bodies, since one body cannot exert a force on another body
unless the other offers a resistance to the one. A force,therefore,
never exists alone. Forces always occur in pairs; one force acting
on one of two bodies and the other force on the other body. In
fact, to every action there is an equal and opposite reaction.
Our conception of force comes mainly from our experiences
in which we have been one of the bodies between which mutual
actions have occurred. The resistance which is offered by bodies
to the action of other bodies arises out of (1) their ability to resist
change of shape (rigidity) and (2) their ability to resist change of
motion (inertia). If a body is acted on by one force, only, a
change of motion of the body will always take place, but if the
body is acted on by two or more forces, it may be held at rest.
Although a single force never exists, it is convenient in the
study of the motion of a body to think of a single force and to
consider only the actions of other bodies on the body in question
'without taking into account the reactions of the body in question
on the other bodies. However, the fundamental nature of force
should hft kent in mind.
4 FUNDAMENTAL CONCEPTIONS AND DEFINITIONS
4. External Effects of a Force.^When a force is applied
to a rigid body the external effect on the body is either to change
the motion of the body acted on or to develop resisting forces
(reactions) between the body acted on and other bodies. Both
of the foregoing effects, of course, may be produced simulta-
neously. For example, consider a body falling freely under the
action of gravity. The sole external effect of the force acting
on the body (its weight) is to produce an acceleration g (32.2 ft.
per sec. 2 approximately). If the same body is placed on the
floor of an elevator which is at rest, the sole external effect of the
weight is to produce an upward reaction of the floor on the body.
If now the elevator moves downward with an acceleration less
than g, the effect of the weight is partly to cause an acceleration
of the body (the same as that of the elevator) and partly to pro-
duce an upward pressure (reaction) of the floor on the body.
The internal effects of a force are to produce stress and defor-
mation in the body on which the force acts. The internal effects
of a force are discussed in books on Strength of Materials.
/- 5. Characteristics of a Force. — From experience we learn that
the external effects of a force depend on, (1) the magnitude of the
force, (2) the position of the Hne of action of the force in the body,
and (3) the sense of the force, that is, the direction along the line of
action. ^ These three properties of a force are called its elements or
characteristics. A change in any one of them causes a change in
the external effect of the force. A discussion of the exact manner
in which these characteristics influence the change of motion of a
body forms an important part of the study of kinetics, and their
influence on the reactions developed in holding a body at rest is
considered in the study of statics.
6. Measure of a Force. Units. — Although we are conscious
of forces of varying magnitudes, we are not able to compare the
magnitudes with precision by means of our muscular sense.
In order to express the magnitude of a force some standard force
must be selected as a unit in terms of which other forces may be
expressed. The unit of force commonly used by the engineer
is the earth-pull (weight) on an arbitrarily chosen body, as found
at a specified or standard position on the earth's surface. Exam-
ples of such units are the pound, ton, kilogram, etc. The earth-
puU on any body (its weight) varies slightly with its position
(altitude and latitude) on the earth. For most engineering cal-
MEASURE OF A FORCE 5
culations, however, the variation in the weight of a body may
be neglected.^
The unit of force as here defined is called a gravitational unit
of force. (For a discussion of other units of force see Arts. 141
and 142.)
There are two common methods used by the engineer for
measuring a force, that is, for finding the number cf units in a
force: (1) by use of the spring balance in its various forms
such as steam gages, certain forms of dynamometers, testing
machines, etc.; (2) by use of a beam or lever balance or system
of levers such as platform scales, screw type of testing ma-
chines, etc.
(1) Due to the fact that many materials possess nearly per-
fect elasticity, within hmits, the unit of force msLj be considered
as the force required to produce a certain stretch or deflection
of an elastic body, the specified stretch or deflection being that
caused by the earth-pull on an arbitrarily chosen body at a stand-
ard locahty. A properly graduated spring balance, therefore,
furnishes one means of measuring any force in terms of the arbi-
trarily chosen unit of force.
(2) In the beam or lever balance, the force to be measured
is applied at one end of a lever or system of levers and an arbi-
trarily selected body is placed at such a position on the other end
of the lever that the earth-pull on the body balances the force.
The arbitrarily selected body is a body on which the earth-pull
at a standard location is the unit of force or some multiple thereof.
The lever is so graduated that the number of units in the unknown
force may be read off directly from the lever or beam.
Attention should here be called to the fact that it is impossible, funda-
mentally, to measure the earth-pull (weight) of a body by means of a beam
or lever balance as stated imder (2) above. The weight of the body, as
already noted, varies with the position of the body on the earth's surface
and this variation is not indicated by the beam balance. Hence if this vari-
ation were large the beam balance could not be used for measuring a force.
Its use, then, is permissible only because the quantity measured (not the
earth-pull) is nearly equal (numerically) to the earth-pull or weight.
* The earth-pull on a body varies directly with g, the acceleration due
to the earth-pull. The extreme variation in the value of g corresponding
to a change in the position of the body on the earth's surface from a high
altitude at the equator to the pole is 0.6 per cent. Within the United States
the maximum variation is about 0.3 per cent.
6 FUNDAMENTAL CONCEPTIONS AND DEFINITIONS
Units of space (length), such as the foot, yard, mile, meter,
etc., and units of time, such as the second, minute, hour, ete., are
assumed to be familiar to the student.
7. Scalar and Vector Quantities. Vector Representation of a
Force. — Quantities which possess magnitude only, as, for example,
areas, volumes, etc., are called scalar quantities. Many quantities
involved in the study of mechanics, however, have direction as
well as magnitude. Any quantity which has direction as an in-
herent property as well as magnitude is called a vector quantity.
Thus, as stated in the preceding article, the effect of a force depends
on its direction as well as its magnitude, and hence force is a vector
quantity. Other examples of vector quantities are velocity, accel-
eration, momentum, etc.
A vector quantity may be conveniently represented wholly or in
part by means of a directed straight Une. Any such line is called a
vector. Thus, the direction of a force may be represented by a
straight line drawn parallel to the action line of the force, the
sense being represented by an arrow-head on the Une, and the
magnitude of the force may be represented by the length of the
line according to some convenient scale. If the magnitude and
direction, only, are to be represented, the vector may be drawn
anywhere in the plane of the force. Such a vector is called a
free vector. If, in addition, the action line of the force is to be
represented, the vector must be laid o€ along the line of action.
Such a vector is called a localized vector. Further, if it is desired
to represent the point of application of the force, the point of
appHcation may be taken as the initial end of the vector, that is,
the point from which the vector is drawn. It will be noted, how-
ever, that the point of apph cation of a force which acts on a rigid
body is not one of the essential characteristics of the force (Art. 5).
In deahng with the forces which act on a given body it is
convenient frequently to represent the forces by free vectors.
The diagram in which are represented the free vectors (that is, the
vectors representing the magnitudes and directions of the forces) is
called the vector diagram.-' The diagram which represents the body
and the action lines of the forces that act on the body is called the
' space diagram. Both diagrams as a rule play an important part
in the complete solution of a problem and will be used frequently
in the subsequent pages./'' In Fig. 1 is shown a wall bracket, the
horizontal arm of which is acted on by three forces having points of
CLASSIFICATION OF FORCES
applications at 1,2, and 3, the action lines being indicated by abj
be, and cd, as shown in the space diagram. The forces are repre-
sented in magnitude and direction in any convenient place by the
vectors AB, BC, and CD, the lengths of the vectors representing
the magnitudes of the forces
according to a convenient scale.
The direction of each vector
is parallel to the action line of
the force which it represents.
The notation used in the
above illustration is known as
Bow's notation and will be used
frequently in the subsequent
pages. According to this nota-
tion the action line of a force is
denoted by two lower-case let-
ters and the vector which re-
presents the magnitude and
direction of the force is denoted
by the corresponding capital
letters.
8. Classification of Forces. Definitions. — Forces may be
classified as surface forces and body forces, sometimes called forces
of contact and forces at a distance according as the action of one
body on another is exerted over a portion of the surfaces of two
bodies that are in contact or is distributed throughout the mate-
rials of the bodies. The most important body force considered in
mechanics is the earth-pull (weight). Magnetic forces are of the
same class. A surface force becomes a concentrated force when
the area over which the force is distributed is so small compared
with the surface of the body acted on that it may be regarded as a
point. This point is called the point of application of the force.
The action line of a concentrated force is a line containing the point
of application of the force and having the same direction as that
of the force.
Any number of forces treated as a group constitute a fwxe
system. A force system is said to be eoneurreni if the action lines
of all the forces intersect in a common point and non-concurrent if
the action lines do not intersect at a point. A force system is said
to be coplanar when all the forces lie in the same plane and non-
FlG. 1.
8 FUNDAMENTAL CONCEPTIONS AND DEFINITIONS
coplanar when the forces do not he in a common plane. A parallel
force system is one in which the action hnes of the forces are par-
allel, the senses of the forces not necessarily being the same, and
a non-parallel system is one in which the action lines of the forces
are not parallel. If the forces of a system have a common line
of action the system is said to be collinear.
Two force systems are said to balance if, when appHed simul-
taneously to a body, they produce no external effect on the body.
The forces which hold a body at rest always balance. Forces or
force systems which balance are said to be in equilibrium and the
body or bodies on which they act are also said to be in equilibrium.
If a body is acted on by a force system which is not in equihbrium
there always is a change in the motion of the body. Such a force
system is said to be unbalanced or to have a resultant. The
resultant of a force system is the simplest equivalent system to
which the given system will reduce, that is, the simplest system
which will produce the same change of motion. The resultant of a
force system is frequently a single force. For some force systems,
however, the simplest equivalent system is composed of two equal,
non-coUinear, parallel forces of opposite sense, called a couple.
And still other force systems reduce to a force and a couple as the
simplest equivalent system. The process of reducing a force
system to a simpler equivalent system is called composition. The
process of expanding a force or a force system into a less simple
equivalent system is called resolution. A component of a force is
one of the two or more forces into which the given force may be
resolved. The anti-resultant or equilibrant of a force system is the
simplest force system which will balance the given system.
9. Principle of Transmissibility. — It was stated in Art. 7
that the external effect of a force on a rigid body does not depend
on the point of application of the force. This very useful fact, the
truth of which is found in our experience, is formally expressed in
the principle of transmissibility. This principle states that the
external effect of a force on a rigid body is the same for all points of
apphcation along its line of action. It will be noted that the
external effect, only, remains unchanged. The internal effect
of a force (stress and deformation) may be greatly influenced by a
change in the point of application along the line of action.
10. Parallelogram Law. — The parallelogram law is the funda-
mental principle on which the composition and the resolution of
PARALLELOGRAM LAW
9
//law
^ of ri
forces is based. The law states tnat the resultant of two forces-
which act on a rigid body is represented in magnitude and in
direction by the diagonal of a parallelogram, the sides of which rep-
resent the magnitudes and the directions of the two forces. In
Fig. 2(a) P and Q are two of the forces acting on a rigid body
(bell-crank), Q being a pressure of 150 lb. at B produced by a force,
P, of 100 lb. Sit A. If Q is laid off from a point 0 (Fig. 26) to some
convenient scale in the proper direction and if, in like manner, P
is laid off from the same
point. then i^^ represents
in magnitude and in di-
rection the resultant of
P and Q. If the point
0 is taken as the inter-
section of the action lines
of P and Q as in Fig. 2a
then the diagonal repre-
sents the action line of
the resultant as well as
its magnitude and direc-
tion.
The parallelogram
is not susceptible
rigid proof. It is an
assumption which ex-
presses a relation be-
tween forces (and other
directed quantities such
as velocities, accelera-
tions, etc.); an assumption, however, which appeals to experi-
ence as reasonable; which is used intuitively in interpreting
and analyzing many of our common experiences ; and which ) ^
thereby becomes a fundamental or basic fact in mechanics.
11. Triangle Law. — The triangle law is a corollary of the par-
allelogram law. It states that the resultant of two concurrent
forces is represented in magnitude and in direction by the third
^ The fact that a force is the resultant of two or more forces is frequently
indicated by two arrow-heads on the vector representing the force. This
fact will also be indicated sometimes by drawing the force vector as a dashed
or broken line.
^//Am////^^^^
Fig. 2.
10 FUNDAMENTAL CONCEPTIONS AND DEFINITIONS
side of a triangle, the other two sides of which represent the two
forces in magnitude and direction. Thus, in Fig. 2(c), the result-
ant R of the two forces P and Q is represented in magnitude and in
direction by the side AC of the triangle ABC in which AB and BC
represent the magnitudes and the directions of QandP respectively.
The action line, ac, of the resultant is parallel to AC and passes
through the point of concurrence of the two forces. 'Although
the triangle law is included in the parallelogram law, its extension
to more than two forces, leading to the force polygon, makes its use
^more convenient than that of the parallelogram law.
Instead of determining the resultant of two forces graphically,
from the parallelogram or the triangle of forces, it may be found
algebraically. Thus, referring to Fig. 2(6), by use of trigonometry,
the magnitude and the direction of the resultant may be expressed
by the equations,
R = VP2_|.Q2_^2PQC0SQ!,
tan 0 = „ , ^ ,
P+Q cos a
where a is the angle between the action lines of Q and P, and 0 is
the angle between the action hues of R and P.
> Although it is not necessary to draw the parallelogram or tri-
angle of forces to scale in determining the resultant of two con-
current forces by the algebraic method, the student should always
make a free-hand sketch of the parallelogram or triangle of forces
when using the above equations.
The expression for the resultant of two forces at right angles
to each other (a = 90°) is of great importance. The resultant
is completely determined by the above equations, which reduce
for this special case to the following equations :
tane = ?
It should be noted that two equations are needed to determme
the magnitude and the direction of a resultant force, whereas one
vector diagram is sufficient for the same purpose.
RESOLUTION OF A FORCE
11
PROBLEMS
1. Two forces having magnitudes of 7 lb. and 3 lb. have the same point of
application on a body. The action line of the 3-lb. force is horizontal with its
sense to the right, while that of the 7-lb. force makes an angle of 45° with the
horizontal with its sense upward to the left. Find the magnitude and the
direction of the resultant. Solve graphically and check the result by the alge-
braic method.
Ans. Magnitude = 5.3 lb.
Direction, 111° 28' with the 3-lb. force.
2. A vertical force of 200 lb. which is appUed at point A of the bell-crank
shown in Fig. 3 causes a horizontal pressure of 300 lb. on the vertical arm at
point B. (a) Find the magnitude
and the direction of the resultant -N (m
of the forces at A and B by the ^>
algebraic method, (b) Find the re-
sultant completely by the graphical
method. 200 lb.
^////////////
Tr'^s y
12. Resolution of a Force.
— In the two preceding arti- ^'^
cles it was assumed that a ^ ._
certain body was acted on by-
two other bodies, and the ac-
tion of a third body was found Fig. 3.
which if allowed to replace
the two would have the same external effect on the body in ques-
tion. The reverse of this process, namely, the resolution of a force,
is also of great importance in mechanics. The action of one body
may be replaced by that of two bodies. The resolution of a force
is acco nplished by means of the parallelogram and triangle laws
and the components (resolved parts) may be found graphically or
algebraically. For example, in Fig. 4(a), Fi represents the steam
pressure on the crosshead, D, of a locomotive and F2 represents the
pressure on the crank-pin, H, of the driver. Let it be required to
resolve Fi into two components, one along the connecting rod
DH, and the other parallel to the crank OH. The action lines, ab
and 6c, of the two components must pass through D as shown in
the space diagram. The magnitudes and the directions of the
components are represented graphically by AB and BC in the
triangle of forces (Fig. 46). This triangle was constructed by
12
FUNDAMENTAL CONCEPTIONS AND DEFINITIONS
laying off AC or Fi parallel to ac and by drawing from A a line
parallel to ab and from C another line parallel to be, the two lines
intersecting at B. Thus, the components of Fi are represented in
magnitude and in direction, but not in action Une, by AB and BC.
The resolution of a force into two rectangular components is of
special importance. The particular value of resolving into rect-
angular components lies in the fact that these components may be
found from very simple algebraic expressions. Thus, let the crank-
pin pressure F2 (Fig. 4a) be resolved into two rectangular com-
FiG. 4.
ponents (Fig. 4c). The action Unes of the components must, of
com-se, pass through the crank-pin, H. Thus, ef is the action hne
of the x-component and fg is the action hne of the ^/-component.
From the triangle of forces the two components EF and FG may
be found. Thus, if </> is the angle between the force and its x-
component, then,
EF = EG cos 0 and FG - EG sin </>.
Or, in general, if F denotes a force which makes an angle d with.
the a;-axis, the x- and ^-components of the force are,
Fx = F cos d and Fy = F sin d
Hence, the eomponent of a foree in a given direction (the other
componeni being perpendicular thereto) equals the product of the
RESOLUTION OF A FORCE
13
magnitude of the force hy the cosine of the acute angle which the ferc&
makes with the given direction.
A force can be resolved into two components in an indefinite
number of ways by drawing lines from any point or pole to the ends
of the force vector. Thus, in Fig. 5 three sets of components are
shown for the force AB, the components being AO and OB for
each position of the pole 0.
It is frequently convenient to resolve a force into three rectan-
gular components. This involves only a slight extension of the
parallelogram law. Thus, the force F (Fig. 6), represented by OA,
Fig. 6.
may be resolved into the two rectangular components OB and OCj
and the component OC may be resolved further into two rectan-
gular components OD and OE. The magnitudes of the com-
ponents of F in the x-, y-, and 2-directions, respectively, are
Fx=F cos Ox Fy = F cos By Fz = F cos dz,
in which 6x, By, and Bz, are the angles which the force makes with
the X-, y-, and 2-directions respectively.
PROBLEMS
3. Given the three concurrent forces as shown in Fig. 7. Find the magni-
tude and the sense of the component of the system (sum of the components
14 FUNDAMENTAL CONCEPTIONS AND DEFINITIONS
of the forces) along the line AB. Solve algebraically and check the result
by the graphical method.
Ans. Magnitude = 24.2 lb.
Sense, Downward to the left.
80 1b
Fig. 7.
Fig. 8.
4. Resolve the weight of 200 lb. (Fig. 8) into (a) two rectangular com-
ponents perpendicular and parallel respectively to AC, (b) into two components
parallel to AC and BC respectively.
5. Resolve the force F (Fig. 9) into three rectangular components in the
X-, y-, and z-directions.
13. Moment of a
Force. — The moment of
a force about (with re-
spect to) a point is the
product of the magni-
tude of the force and
the perpendicular dis-
tance from the point to
the action Une of the force.
Thus, the moment of
the force /^ (Fig. 10) about
the point 0 is Fd. The
Y point about which the
Fig. 9. moment is taken is called
the moment-center (or
origin) and the distance d is called the moment-arm. A unit \
moment is the product of a unit force and a unit distance such
z
<py
<'/
.f
/
/ /
/
/
0
/
/.
/
/
/
MOMENT OF A FORCE
15
Fig. 10.
as a Ib.-ft., Ib.-in., ton-ft., etc. The moment of a force witlr^
respect to a point may also be regarded as the moment of the
force with respect to the line which passes through the point
and is perpendicular to the plane determined by the point and
the action line of the
"^ force. Thus, in Fig. 10,
Fd also represents the
moment of F about the
axis YY.
The physical signifi-
cance of the moment
of a force as defined
above is built up in-
tuitively from our ex-
periences in which we
have exerted moments
on other bodies. It ex-
presses a measure of the tendency of the force to turn the body ,
on which it acts , about a given axis. ^- Strictly speaking, the
fact that the tendency of a force to turn a body about a given
axis is directly proportional to the moment of the force, is an
assumption which, when reinforced by experience and the anal-
ysis of observed facts,
crystallizes into a fun-
damental or basic truth
of mechanics.
In general, the mo-
ment of a force about
an axis which is not
perpendicular to the
plane of the force is de-
fined as the moment of
the component of the
force perpendicular to
the given axis, the other
component of the force being parallel to the axis. Thus, in
Fig. 11 the moment of the force F about the axis ZZ is FyOA.
As a rule it will be convenient to select the moment axis as
one of the coordinate axes. The moment of a force with respect
Fig. 11.
16
FUNDAMENTAL CONCEPTIONS AND DEFINITIONS
to a coordinate axis will be considered as positive if the direction
of rotation is counter-clockwise when viewed from the positive end
of the axis.
PROBLEMS
6. A force of 20 lb. is exerted on the knob of a door as shown in Fig. 12.
If the action line of the force lies in a plane perpendicular to the door, what is
the moment of the force about the axis YY? Ans. 42 A lb .-ft.
z
/
/
/
/
<
/
—
y
/
^
/
/
V
/
Fig. 12.
Fig. 13.
7. Find the moment of the 40-lb. force (Fig. 13) with respect to each of the
coordinate axes; each division represents 1 in.
14. Principle of Moments. Varignon's Theorem. — The prin-
ciple of moments is of great importance in mechanics. It appUes
to lines, areas, volumes, etc., as well as to forces. It will be con-
sidered, however, at this point only in connection with two con-
current forces. ^ The principle for this restricted case, which is
known as Varignon's theorem, states that the algebraic sum of the
moments of two concurrent forces about any point in their plane is
equal to the moment of their resultant about the same point.
V The principle of moments, Hke the parallelogram law, is an
assumption; an assumption, however, which appeals to experi-
ence as reasonable; which is frequently and intuitively used in
interpreting and analj^zing observed facts, under various con-
ditions; and which thereby becomes one of the fundamental
facts or principles of mechanics.
PRINCIPLE OF MOMENTS
17
The fact that the principle of moments for forces is in agree-_
ment with the parallelogram law may be shown by deducing the
principle from the parallelogram law as follows: In Fig. 14 let
P and Q represent two forces concurrent at A, the resultant accord-
ing to the parallelogram
law being R. Let 0 be
any moment-center in
the plane of the forces.
It is required to prove
that
where p, q, and r are
the moment-arms of P,
Q, and R, respectively. jtjg^ 14,
Let a set of rectangular
coordinate axes AX and AY be chosen as shown in the figure,
A Y passing through the moment-center 0. Let a, j8, and 6
denote the angles which the action Unes of P, Q, and R, respect-
ively, make with the AX axis. From the figure it is seen that
that is,
FG-\:AF = AG,
P cos a+Q cos fi = R cos 6.
By multiplying both sides of this equation by AO, the following
equation is obtained :
Hence,
P'AO cos a+Q'AO cos ^ = R'AO cos 0.
Pp+Qq = Rr.
It will be noted that the definition of the moment of a force
about a line as given in Art. 13 is in accordance with the principle
of moments. It is often convenient to obtain the moment of a
force about a point in its plane (or about an axis through the point
perpendicular to the plane) by finding the sum of the moments of
its rectangular components. In accordance with the principle of
transmissibility a force may be resolved into components at any
point along its line of action. Thus, in Fig. 15, Fx and Fy are one
18
FUNDAMENTAL CONCEPTIONS AND DEFINITIONS
pair of rectangular components and F'x and F'y are another pair
of components of the force F. The moment of the force F about
the axis ZZ is, by the principle of moments, equal to the sum of
Fig. 15.
the moments of the forces of either pair of components. Hence,
the moment of F may be expressed as follows :
Fd = Fydx-F:c-dy=^F',-d'y+F'yd\.
If the force F is not in a plane perpendicular to the axis, as in
Fig. 16, the force may
be resolved into three
rectangular compo-
nents, one being par-
allel to the axis and
the other two in a
plane which is perpen-
dicular to the axis.
The moment of the
force is then the sum
of the moments of the
components which lie
in this plane. Thus,
Fig. 16. • r^- -ia 4.x.
in Fig. 16, the mo-
ment of F about the axis MN is,
-Fx'dy + Fy'd,.
The component Fz, being parallel to the axis MN, has no moment
about the line MN.
r
COUPLES
19
Fig. 17.
15. Couples. — Two equal parallel forces which are opposite ia-
sense and are not collinear are called a couple. A couple cannot
be reduced to any simpler force system. The fact that the only
effect of a couple is to produce or to prevent turning is obtained
intuitively. The moment of a couple about any point in the
plane of the couple (or any axis perpendicular to the plane of the
couple) is defined as
the algebraic sum of
the moments of the
forces of the couple
about the point (or
axis). From this defi-
nition it follows that
the moment of a couple
about any point in
the plane of the couple
(or axis perpendicular
to its plane) is the
product of the magni-
tude of either force of
the couple and the perpendicular distance (moment-arm) between
the action fines of the two forces. This statement may be
proved as foUows: In Fig. 17 let F, P be the two forces of a
couple and 0 any point in their plane or, YY any axis perpen-
dicular to their plane. The algebraic sum of the moments of
the two forces about 0 (or YY) is
P'OB-P'OA,
which may be written thus,
P{()B-Ol)=P'AB = Pp.
In fike manner the moment of the couple may be shown to be the
same about any other point in the plane (or any other axis parallel
to YY). Since the moment of a couple depends only on the
product of either force of the couple and the arm of the couple it
follows that the turning effect of a couple on a rigid body about an
axis in the body is the same for different magnitudes and fines of
action of the forces, provided that the moment of the couple
remains constant. For example, in Fig. 18, if cords are wrapped
20
FUNDAMENTAL CONCEPTIONS AND DEFINITIONS
around two pulleys, of radii ri and r2, which are keyed together
and equal weights are attached to the other ends of the cords, the
pulleys will rotate exactly the same as they would if forces F, F
were applied as shown by the dotted Unes, provided that the
moments of the two couples are the same, that is if W{r2—ri) is
equal to F-2r2. On the other
hand if motion is to be prevented
a couple having a moment of
— {F ' 2r2) would be required.
16. Characteristics of a
Couple. — The external effect of
a couple when appUed to a rigid
body is either to cause a change in
the rotational motion of the body
or to develop a resisting couple
due to the actions of other bodies
on the body in question. Both of
the foregoing effects may of course
be produced simultaneously.
From experience we learn that
the external effect of a couple
depends on (1) the magnitude
of the moment of the couple, (2)
the sense or direction of rotation of the couple, and (3) the aspect
of the plane of the couple, that is, the direction or slope of the plane
(not its location) . These three properties of a couple are called its
characteristics. Since parallel planes have the same aspect it
follows from what has been stated above that two couples which
have the same moment and sense are equivalent if their planes are
parallel. The fact that the external effect of a couple is independ-
ent of the position of the plane of the couple and depends only
on the direction of the plane is amply verified by experience.
7 Thus, in screwing a pipe into a joint by means of two pipe wrenches
/ the forces applied at the end of the wrenches constitute a couple,
y and it is a matter of experience that the effort required is the same
\ regardless of the position along the pipe at which the wrenches are
applied (as^ming perfect ahgnment between pipe and joint).
17. Vector Representation of a Couple. — In dealing with
couples it is convenient to represent the couples by means of vec-
tors. In order to represent a couple completely by a vector all of
Fig. 18.
VECTOR REPRESENTATION OF A COUPLE
21
the characteristics of the couple must be represented. The moment"
of the couple may be represented to scale by the length of the vec-
tor. The aspect of the couple may be represented by drawing the
vector perpendicular to the plane of
the couple and hence parallel to the
axis of the couple — that is, parallel
to the axis about which the couple
rotates or tends to rotate the body
on which it acts. The sense of the
couple may be represented by an
arrow-head on the vector, the usual
convention being to direct the arrow-
head away from the plane of the
couple in the direction from which the
rotation appears counter-clockwise.
This method of representing the sense
of a couple involves the so-called con-
vention of the right-handed screw,
since a right-handed screw having an
axis perpendicular to the plane of the couple would move in the
direction of the arrow if given a rotation which agrees in sense
with that of the couple. Thus, in Fig. 19 the vector OA com-
pletely represents the couple Fd, provided that the length of OA
represents to scale the product F • d.
Fig. 19.
PROBLEMS
8. The steering wheel of an automobile shown in Fig. 20 is 18 in. in diam-
eter. Forces exerted by the hands on the wheel constitute a couple which is
Fig. 21.
22 FUNDAMENTAL CONCEPTIONS AND DEFINITIONS
Fig. 22.
represented by the vector AB. If the length of ^B is 1.5 in. and the scale
used is 1 in. =60 lb .-in., represent the forces of the couple on the circumference
of the wheel.
9. A couple having a moment of 60 lb. -in. is required to open the blow-off
valve shown in Fig, 21. Represent the couple completely by a vector, using a
scale of 1 in. =24 Ib.-in.
18. Resolution of a Force into a Force and a Couple. — In many
problems, both in statics and kinetics, it is convenient to resolve a
force into a force parallel to the
given force, and a couple in
the plane of the force. Thus, in
Fig. 22, let P represent a force
acting on a body at A. Let
two equal, opposite, and col-
linear forces P, P be introduced
at any point 0, each of the
forces being parallel to the orig-
inal force and of the same
magnitude. The three forces
are equivalent to the original one, since the two equal, opposite,
and collinear forces have no effect on the body. The force system
may now be considered to be a force P acting at 0 (parallel to the
given force and of the same magnitude and sense), and a couple,
the moment of which is the same as the moment of the original
force about 0.
The action lines of the forces of the couple, however, may be
moved to any location in the plane of the forces and the magnitude
of the forces of the couple may be changed to any value, pro-
vided that the forces remain equal and parallel to each other and
also that the moment and sense of the couple are not changed.
Couples will be discussed in greater detail in Art. 27.
wY Since a force may be resolved into a force _an(i couple in the
M plane 'of tlie force it follows that a force and a couple in a plane may
be combined into a resultant force in the same plane. Further,
the resultant force has the same magnitude, direction, and sense
as the given force and its action line is parallel to the action line
of the given force. In other words, the sole effect of combining a
couple with a force is to move the action line of the force into
a parallel position, the force being unchanged in all other
respects.
RESOLUTION OF FORCE INTO FORCE AND COUPLE
23
ILLUSTRATIVE PROBLEM
10. In Fig. 23 is shown an arm mounted on an axle 0. A vertical force
P = 20 lb. acts on the arm at C. Resolve the force P into a force acting
through 0 and a couple the forces of which act at A and B.
P= 20 lb.
40 1b.
40 1b.
■>*
< 10^^ — ^
20 1b.
0
^r=3
20 1b.
Fig. 23.
Solution. — By introducing two vertical 20-lb. forces at 0 which are oppo-
site in sense the force is resolved into a vertical downward force of 20 lb. at O
and a couple having a moment of 20X10 = 200 lb .-in. This couple may be
replaced by two horizontal forces, of 40 lb. each, acting at A and B as shown.
The force at O and the couple will produce exactly the same external effect
on the arm as the original force at C.
PROBLEMS
11. A gusset plate B (Fig. 24) is attached to another plate A by means of
four rivets as shown. A force of 10,000 lb. at the point 0 (1.5 in. below
dotted line) is transmitted to the member A. Resolve this force into a force,
acting along the center line of the rivets, and a couple.
Fig. 24.
^W=20 lb.
Fig. 25.
12. A body weighing 20 lb. is mounted on an axle 0 (Fig. 25) and is acted
on by a couple as shown in the figure, in addition to its weight and the axle
reaction at 0. Replace the weight and the couple by an equivalent single
force.
24 FUNDAMENTAL CONCEPTIONS AND DEFINITIONS
Fig. 26.
13. A load P of 100 lb. acting on a wall bracket as shown in Fig. 26 causes
the wall to exert a horizontal force of 150 lb. and a vertical force of 50 lb. at
each of the wall attachments,
p-iooib. Consider the two 50-lb. forces
to be equivalent to one 100-lb.
force and replace this 100-lb
force and the two 1 50-lb.
forces by an equivalent single
force.
19. Methods of Solution
of Problems. A Igebraic and
Graphical Methods. — In the
analysis and solution of
problems in mechanics two
general methods may be
used, namely, the algebraic
and the graphical methods.
In the algebraic method of solution, quantities are represented
by letters or by numbers which are dealt with according to
the rules of mathematics. In the graphical method of solu-
tion, quantities are represented by lines which are dealt with
by means of geometric constructions. Simple graphical methods
have already been used in the preceding articles in con-
nection with forces, couples, etc. In general, either of the two
methods may be used in the solution of a problem. Some prob-
lems, however, may be solved much more easily by the algebraic
method while other problems yield much more readily to the
graphical method. The operations involved in the solution of a
problem by the two methods are so radically different that one
method of solution serves as an excellent check on the other
method.
Approximate Methods. — In making computations it is impor-
tant to keep in mind the degree of accuracy which should be
obtained. The degree of accuracy desired will depend, in general,
on two factors, namely:
(1) The degree of accuracy of the original data or quantities
on which the computation is based, and
(2) The use which is to be made of the computed results.
The data on which many engineering computations are based
are determined from experiments and hence are approximate
values, the degree of approximation depending on the instruments
METHODS OF SOLUTION 25
and methods used, and on the care and skill of the observer, ^lie
computed results which are based on these values cannot have a
greater degree of accuracy than that of the original data. In
general a sum, difference, product, or quotient of two approximate
values will not have a greater degree of accuracy than that of
the least accurate of the two numbers. For example, if one
numerical quantity is accurate to two significant figures and
another quantity to three significant figures, the product of the
two numbers will not be accurate to more than two significant
figures, and hence more than two significant figures should not be
retained in the result.
Dimensional Equations. — In algebraic equations in which the
variables represent physical quantities, all of the terms of the
equation must be of the same dimensions, or, to express the same
idea in mathematical language, an algebraic equation which
expresses a relation between physical quantities must be homo-
geneous. The use of this principle is of assistance in checking any
equation for correctness and in determining the specific units in
which a result is expressed when computed from a given equation.
For each of these purposes the given algebraic equation is replaced
by a dimensional equation.
The dimensional equation corresponding to any algebraic
equation is formed by replacing each term of the given equation
by a term which indicates the kinds of fundamental quantities
in wkich the term is expressed and which also indicates the degree
of the corresponding quantities in each term.
The fundamental quantities used in engineering are force,
length, and time {F, L, and T). Hence, in an equation, a term
which represents an area is replaced by L^ in the dimensional
equation since an area is the square of a length. A velocity is a
length divided by time and hence is represented in the dimensional
equation by LT~^ and similarly for other quantities. It should
be noted that in the dimensional equation only the kinds of funda-
mental quantities are indicated and not the specific units used in
measuring these quantities and also that the number of such units
is not indicated. Hence, numerical constants in the algebraic
equation do not appear in the dimensional equation.
For example, consider the equation ad?-\-d^ = v, in which a
represents an area, d a length, and v a volume. Since an area is
the square of a length (L^) and a volume is the cube of a length
26 FUNDAMENTAL CONCEPTIONS AND DEFINITIONS
(L^), the dimensional equation is V'-\-L^ = L^ and hence the given
equation is incorrect.
Consider also the equation P-\-kv = as in which P represents a
force, k a weight (force) per unic volume, v a volume, a an area, and
s a force per unit area. The dimensional equation then may be
written,
That is, F-\-F = F,
and hence the equation is dimensionally correct.
PI
Further, consider the equation E = — , in which P represents a
ae
force in pounds, I a length in inches, a an area in square inches,
and e a length in inches. Let it be required to determine the units
in "which E is expressed. The dimensional equation is
L^XL L2'
Hence, in accordance with the units stated (pound and inch), E
is expressed in pounds per square inch (Ib./sq. in. or Ib./in.^).
PROBLEMS
14. The equation Q = .622 ■\/2^ib—0.2h) h^ is used to determine the quan-
tity of water flowing per unit of time over a rectangular weir of width b when
the height of water above the weir is h, g being the acceleration due to gravity
(32.2 ft. /sec. 2). (a) Write the dimensional equation. (6) What are the units
of Q if 6 and h are expressed in feet? (c) If Q equals 24.4 cu. ft. per sec. and b
equals 3 ft. compute the value of h to the same degree of accuracy as is used in
expressing Q, that is, to three significant figures. Note: This formula is
valid only when .2h is small compared with b. Hence the term .2h may be
omitted and a trial value of h may be found by solving the resulting equation.
This value of h may now be used in the term .2h and a second trial value of h
may be obtained by solving the resulting equation. This process may be
repeated.
15. In finding the diameter of a pipe required to discharge a given quantity
of water, the formula d^ = Ad-{-B is used, in which A and B are known quan-
tities. If A = 15.5 and J5=400 compute the value of d to three significant
figures. If the given equation is dimensionally correct what can be said of
the quantities A and 5?
DIMENSIONAL EQUATIONS 27
16. In the following equation, T is the moment of a force, *S is a force per
unit of area in pounds per square inch, A, ho, h, and bg, are lengths in
inches. Is the equation dimensionally correct? What are the units of T?
_ 2 b'h-bo\
CHAPTER II
RESULTANTS OF FORCE SYSTEMS
20. Introduction. — The determination of the resultants of
various force systems as discussed in this chapter is of importance,
mainly, (1) in the study of the conditions which the forces of a
system satisfy when they hold a body in equiUbrium (Statics,
Chapter III), and (2) in the study of the laws by which the motions
of bodies are governed (Part HI, Kinetics).
(1) The equations of equiUbrium for a given type of force
system are obtained by expressing the conditions which the forces
must satisfy in order that the resultant of the system shall be
zero. Therefore the resultant to which a given type of force
system reduces must be known before the conditions which are
required to make the resultant equal to zero can be established.
Furthermore, in dealing with forces in equilibrium it is frequently
convenient to replace several of the forces of a balanced system
by the resultant of the several forces and to deal with the result-
ing force system instead of the original system.
(2) The motions of bodies are influenced by the unbalanced
part (resultant) of the forces which act on the bodies. In the
study of the motions of physical bodies, therefore, a knowledge of
the resultants of various force systems and of methods of expressing
the characteristics of resultants in terms of the forces of the system
must be understood.
§ 1. CoLLiNEAR Forces
21. Algebraic Method. — The resultant will be found first for
two collinear forces, P and Q, having the same sense. It is a matter
of expei-ience that the two forces are equivalent to a single force
which has a magnitude equal to P-\-Q and a line of action and
sense which are the same as the line of action and sense of the
given forces. This proposition may be proved, however, by apply-
ing the parallelogram law. Thus, according to this law, the result-
ant of any two concurrent forces P and Q, the lines of action of
28
GRAPHICAL METHODS 29
which make an angle a with each other, is a single force of magni-
nitude, R, which is given by the equation R =^ P'^-\-Q^-\-2PQ cos a
(Art. 11). In the special case of two coUinear forces here consid-
ered, a equals zero and hence i2 = Vp2_|_Q2^2PQ = P+Q. Sim-
ilarly, the resultant of two forces P and Q having opposite senses
(P being the larger of the two forces) is a force the magnitude of
which is given by the equation R = P—Q, the sense of R being the
same as the sense of the larger force P. Hence, the resultant of
any two collinear forces is a single force having the same Hne of
action as the given forces, the magnitude and sense being indicated
by the algebraic sum of the forces. The extension of this method
to any number of colhnear forces may easily be made. Thus the
resultant of two of the forces may be combined with a third force;
the resultant thus obtained may then be combined with a fourth
force, and so on, until the entire system is reduced to a single
resultant force. Therefore the magnitude of the resultant of a
collinear force system is given by the equation,
R=J1F.
§ 2. Concurrent Forces in a Plane
22. Graphical Methods. First Method. — The resultant of an
unbalanced system of concurrent forces in a plane is a force which
may be found by means of the parallelogram law. In Fig. 27
are shown three forces Pi, P2, and
Fs which act on a body at the point
0. The forces Pi and P2 may be
combined into a single force Pi by
means of the parallelogram law.
Similarly Pi and P3 rnay be com-
bined into a single force P2. P2
then is the resultant of the given
forces. By continuing this process Fig. 27.
any number of concurrent forces
may be combined into a single force. The order in which the
forces are taken is immaterial. If the resultant force obtained by
combining all except one of the forces of a concurrent system is
equal to that one, coUinear with it, and of opposite sense, the two
30
RESULTANTS OF FORCE SYSTEMS
forces cancel and hence the resultant of the given system is
equal to zero.
Second Method. — Another graphical method of determining the
resultant of a system of concurrent forces in a plane involves the
apphcation of the triangle law. Consider, for example, the three
forces Fi, F2, and F2,, which act on a body and concur at a point 0
as shown in Fig. 28(a). In order to determine the resultant of
the three forces, draw from any arbitrary point A (Fig. 286), a
vector representing the magnitude and the direction of the force Fi ;
from the end of this vector draw another vector representing the
(a)
Fig. 28.
magnitude and the direction of the force F2. Ri, the resultant of
Fi and F2, is represented in direction and magnitude by the vector
drawn from A to the end of F2. To find the resultant of i^i and F3,
and hence of the three given forces, draw from the end of Ri
(or F2) a vector representing Fz in magnitude and in direction.
The resultant of R\ and F3 is then represented in magnitude and in
direction by the vector R2 drawn from A to the end of F3. It
should be noted that this vector R2 represents the magnitude and
the direction, only, of the resultant of the given forces and not the
action line of the resultant. The action line of the resultant must
pass through the point 0 in the body at which the forces are con-
current.
This method of determining the resultant may be stated
formally as follows: In order to find the resultant of a system of
concurrent forces in a plane, construct a polygon the sides of
which are vectors representing the given forces in magnitude and in
direction; a Une drawn from the beginning of the first vector to
ALGEBRAIC METHOD
31
the end of the last vector represents the magnitude and the direc-
tion of the resultant of the given system. If the force polygon
closes, that is, if the end of the last vector coincides with the begin-
ning of the first vector, the resultant of the given system is equal
to zero.
23. Algebraic Method. — In Fig. 29(a) is represented a coplanar
force system which is concurrent. Each of the four forces of the
system may be resolved into two components, one lying along the
a;-axis and one along the ^/-axis. The a;-components of the forces
constitute a coUinear system the resultant of which is a force along
-
^
R
y^
J>
u.
/
H
./^
2Fc
(&)
(a)
Fig. 29
the a;-axis. The magnitude of this resultant is equal to SFx. The
^/-components of the forces Ukewise constitute a coUinear system
the resultant of which is a force along the 2/-axis of magnitude
'ZFy. Since the system, by this method, is reduced to two forces
the resultant of the given system is the resultant of these two forces.
If the magnitude of this resultant is denoted by R and the direction
which its action line makes with the a:-axis by the angle dx as shown
in Fig. 29(6), the resultant may be found from the equations:
i? = V(SF,)2+(SF^)2,
tane,
The action line of the resultant must of course pass through the
point of concurrence of th forces.
32
RESULTANTS OF FORCE SYSTEMS
ILLUSTRATIVE PROBLEM
17. Find the resultant of the system of concurrent forces shown in Fig
30(o).
Algebraic Solution:
XFx = 20 cos 30° -30 cos 60° -10 cos 45° +25 cos 45°
= 17.32-15-7.07 + 17.67
= 12.92 lb.
7:Fy = 20 cos 60° +30 cos 30° - 10 cos 45° - 25 cos 45°
= 10+25.98-7.07-17.67
= 11.24 lb.
and
R = V(12.92)2+(11.24)2 = 17.1 lb. (Fig. 306)
^-*^^-^^' = ^^'
W S Fa.=12.92 lb.
(6)
Fig. 30.
(c)
Graphical Solution:
To determine the resultant of the system graphically a force polygon is
constructed as shown in Fig. 30(c) . The resultant is represented by the vector
AE. The magnitude of the resultant is found, by measuring the length of
AE, to be 17.1 lb. and the angle which the resultant makes with the a;-axis is
found to be 41°.
PROBLEMS
In the following three problems the forces are concurrent at the origin.
The values of F are the magnitudes of the forces and the values of Bx are the
angles which the action lines of the forces make with the positive end of the
X-axis, the angles being measured in a counter-clockwise direction. In each
problem it is required to find the resultant of the forces specified
18.
19.
20.
ALGEBRAIC METHOD
F 1001b. 601b. 40 1b. 50 1b.
dx 30° 135° 240° 330°
Ans. 72 = 75 lb., (?x = 25° 55'.
F 20 1b. 10 1b. 15 1b. 5 1b.
dx 0° 45° 120° 270°
Ans. 72=24.7 lb., 0^=37° 35'.
F 25 1b. 101b. 301b. 401b.
33
Ox 0° 30'
135° 240°
21. Fig. 31 represents a body acted on by four forces. Determine com-
pletely the resultant of the four forces by the graphical method and check the
results by the algebraic method. Ans. 72 = 186 lb.; ^a; = 164° 45'.
100 lb.
Fig. 31.
Fig. 32.
22. The body represented
in Fig. 32 is acted on by five
forces as shown. Replace the
two 80-lb. forces by a single
force and then find the resultant
of this force and the remaining
three forces.
23. Find by the graphical
method the resultant of the five
forces shown in Fig. 33.
24. Solve the preceding
problem by the algebraic
method.
^^
f6
■^
\ -
\
\
/
\
y
10 lb._
/
/
£
r
Fig. 33.
34
RESULTANTS OF FORCE SYSTEMS
§3. Parallel Forces in a Plane
24. Graphical Methods. First Method. — A system of two
parallel forces having the same sense will first be considered.
In Fig. 34 are shown two such forces, P and Q, the points of appU-
cation of which are B and A respectively. The resultant of these
two forces cannot be found directly by the parallelogram law. How-
ever, by introducing at B and A any two equal, opposite, and colli-
near forces F, F, as shown in Fig. 34, the parallelogram law may be
used. Thus the forces
Q and F may be re-
placed by their result-
ant Ri. Likewise P
and F may be replaced
by their resultant R2.
At C, the point of in-
tersection of the Hues
of action of Ri and R2,
let Ri be resolved into
components Q and F,
and R2 into compo-
nents P and F. The
two forces F, F at C
cancel, thereby reduc-
ing the system to two colhnear forces P and Q along the hne DC.
The resultant of these two forces, which is also the resultant of
the original two forces, is a single force of magnitude P-\-Q
along the hne DC. In order to determine the position of the
action Hne DC of the resultant, consider the two similar triangles
AJG and ACD. From these similar triangles there is obtained
the relation.
Fig. 34
Q_CD
F AD
(1)
Similarly, from the triangles BKH and BCD the following rela-
tion is obtained,
P CD
F BD
(2)
GRAPHICAL METHODS 35
By eliminating F from equations (1) and (2) the resulting equation"
is
Q BD
Hence the resultant of two parallel forces of like sense is a
force having a line of action which is parallel to the lines of action
of the two forces and which divides the distance between them
internally into segments which are inversely proportional to
the magnitudes of the forces. The magnitude of the resultant is
the sum of the forces and its sense is the same as that of the forces.
If the forces P and Q are unlike in sense it can be shown in a
manner similar to that used above that their resultant is a force
of magnitude P—Q(P being larger than Q), the sense of which is
the same as that of P. The line of action of the resultant is parallel
to the lines of action of the forces and divides the distance between
them externally into segments which are inversely proportional to
the magnitudes of the forces. The proof is left to the student.
It should be noted that the construction used in Fig. 34 fails
if the two opposite, parallel forces are equal in magnitude since two
such forces constitute a couple and cannot be replaced by a single
force.
This method of finding the resultant of two parallel forces can
be extended to any number of parallel forces. For, any two forces
of such a system can be combined into a single resultant force;
this resultant and a third force of the system can be combined in
exactly the same manner, and so on, until the entire system has.
been reduced to a single resultant force. It is evident that the
magnitude of the resultant is equal to the algebraic sum of the
given forces. If in combining the forces of a parallel system by the
above method it is found that the resultant of all except one of the
forces of the system is a force which is equal to the force omitted
and opposite to it in sense, the two forces constitute a couple
which is the resultant of the system. If, however, the lines of
action of the two forces just mentioned are collinear, the forces
balance and the resultant of the system is equal to zero. Hence
if the resultant of a system of parallel forces in a plane is not
equal to zero, the resultant is either a force or a couple.
The determination of the line of action of the resultant force
by the method just discussed involves considerable work if there
36
RESULTANTS OF FORCE SYSTEMS
are many forces in the system. A second graphical method will
now be considered. This method involves less construction than
the method just considered and possesses other important advan-
tages.
Second Method. — In Fig. 35(a) is shown a system of four par-
allel forces having the action Hnes ab, he, cd, and de. On a Hne par-
allel to the action lines of the forces are laid off vectors AB, BC,
CD, and DE which represent the given forces in magnitude and
direction (Fig. 356). From any arbitrarily chosen point, 0, Unes
are drawn to the points A, B, C, D, and E. The figure thus con-
structed is called a force polygon; the point 0 is called the pole,
^
^'^o
^""b
n C
(a)
""^-.
— >- — .^
%^
Fig. 35.
and the lines OA, OB, OC, OD, and OE are called rays. From the
triangle law it follows that by this construction the force AB is
resolved into two components represented in magnitude and in
direction by the rays AO and OB. AO and OB, however, do not
represent the Unes of action of the two components, since the lines of
action of the components of AB must intersect on ab. Similarly,
BO and OC represent the magnitudes and the directions of two
components into which BC is resolved, and so on for the remaining
rays. The given system then may be replaced by another system
of eight forces which are represented in direction and in magni-
tude by the rays AO, OB, BO, OC, CO, etc. The forces repre-
sented by the rays OB and BO (OC and CO, etc.), are equal in
magnitude, opposite in sense, and their lines of action are parallel.
GRAPHICAL METHODS 37
If now the action lines of the pairs of parallel and equal forces 0B~~^
and BO, OC and CO, OD and DO, are made to coincide, each pair
will cancel and there will remain only two forces, AO and OE.
The resultant of these two forces is the resultant 'of the given
system and is represented in magnitude and in direction by the
vector AE.
The action line of AE will pass through the point of intersection
of the action hues of its components AO and OE. In order to locate
the point of intersection of the action Hnes of these two forces it is
necessary to construct another figure called the funicular polygon.
This is a polygon the sides of which, called strings, are parallel to
the rays of the force polygon. The funicular polygon is con-
structed as follows: From any point on ah, the action fine of AB,
strings ao and oh are drawn parallel to the rays AO and OB.
These strings represent the action lines of the forces AO and OB
into which AB was resolved. From the point of intersection of oh
and he the string ho (which wiU coincide with the string oh) and
the string oc are drawn. From the point of intersection of oc
and cd are drawn co and od and from the point of intersection of od
and de are drawn do and oe. Since oh and ho, the action lines of
the equal and opposite forces OB and BO, are coUinear, the two
forces OB and BO will cancel. Similarly for OC and CO and for
OD and DO. The system then is reduced to two forces repre-
sented in magnitude and in direction by the vectors AO and OE,
the action hnes of the forces being ao and oe. The resultant of
these two forces, which is also the resultant of the given system,
will pass through the point in which their action lines ao and oe
intersect. The resultant then is completely determined by the
force and funicular polygons; the former determines its magni-
tude and direction; the latter determines a point on its action fine.
If the force polygon closes, that is, if E coincides with A, the
rays AO and OE represent two forces which are equal in magni-
tude and which have parallel action hnes and opposite senses.
The resultant in this case is a couple provided that the fimicular
polygon does not close. If, however, the funicular polygon also
closes, that is if ao and oe coincide, the two forces AO and OE wiU
balance and the resultant of the given system will be equal to
zero.
In constructing the force and funicular polygons any point may
be taken for the pole and the funicular polygon may be started
38
RESULTANTS OF FORCE SYSTEMS
at any point on the action line of any one of the forces. A change
in the positions of these points has the effect, only, of locating a
different point on the action line of the resultant if the resultant of
the system is a force. If the resultant of the system is a couple,
the effect is to change both the magnitude of the forces constituting
the resultant couple and the length of its moment-arm. The
couples will be equivalent, however; that is, they will have the
same moment and sense.
ILLUSTRATIVE PROBLEM
26. Find by the graphical method the resultant of the four parallel forces
shown in Fig. 36(a).
10 1b.
20 1b.
25 lb. 15 lb.
b
6
a ^
c c
f J
d d
^-. '
0 ^
o^^^
^^^c
<- ^
No
(c
^f '
E,A
N
--^
Scales; lin.= 2ft.
W
l.m.= 20 1b.^
Fig, 36.
Solution. — Since the force polygon closes the resultant is not a force and
hence is a couple. In constructing the rays (Fig. 366) it is convenient to take
the pole 0 so that the force represented by OA has an integral value. OA here
represents to scale 20 lb. The funicular polygon in Fig. 36(a) is constructed
according to the method described in Art. 24. The resultant is a couple
consisting of the two 20-lb. forces AO and OE. The arm as scaled from the
funicular polygon is 1.25 ft. Hence the moment of the resultant couple is
— 25 lb. -ft., the minus sign indicating a clockwise direction of rotation.
25. Principle of Moments. — The principle of moments as dis-
cussed in Art. 14 for the special case of two concurrent forces may
be extended to all force systems. Briefly, the principle states that
the moment of the resultant of a force system is equal to the alge-
braic sum of the moments of the forces of the system. The prin-
ciple is of great importance in the determination, by the algebraic
method, of (a) the action line of the resultant of a system of forces
THE PRINCIPLE OF MOMENTS 39
when the resultant is a force, and (6) the moment and sense of the^
resultant of the force system when the resultant is a couple. A
formal statement and proof of the principle of moments will not
be given for each of the force systems considered since the method
of proof is substantially the same for all the force systems. As
applying to a system of parallel forces in a plane the principle may
be stated formally as follows : The algebraic sum of the moments of
the farces of a coplanar parallel system about any point in their plane
is equal to the moment of the resultant of the system about the same
point.
In demonstrating the truth of this statement use will be made of
the diagram in Fig. 35 and of the methods of the preceding article.
The given system of four forces (Fig. 35) was replaced by another
system of eight forces which were represented in magnitude and
in direction by the rays of the force polygon and in line of action
by the strings of the funicular polygon. Six of these forces occur
in pairs, the two forces of each pair being coUinear, equal in magni-
tude, and opposite in sense. Obviously, the sum of the moments
of the two forces of each pair with respect to any point in their
plane is equal to zero. For any moment-center in the plane, by
use of Varignon's theorem, the following relations may be written,
moment of AB = moment of AO+moment of OB,
BC= " B0+ " OC,
CD= " C0-\- " OD,
DE= '' D0+ '' OE.
If the two sides of the above equations are added the result
obtained may be stated as follows:
The sum of the moments of the forces of the system
= moment of AO+moment of OE,
since the remaining terms on the right side of the equations cancel
in pairs. But AO and OE are the components of the resultant
force of the system and hence, by Varignon's theorem, the sum of
the moments of AO and OE is equal to the moment of the resultant
of the system. Hence, the proposition is proved for a parallel
force system in which the resultant is a force.
If the resultant of the force system is a couple, that is, if the
forces AO and OE are parallel, equal, and opposite as in Fig. 36,
40
RESULTANTS OF FORCE SYSTEMS
the proposition also holds, since the sum of the moments of AO
and OE is equal to the moment of the resultant couple.
26. Algebraic Method. — A system of coplanar parallel forces
is shown in Fig. 37, the forces being parallel to the y-axis. In the
graphical determination of the resultant of such a system it was
seen that the resultant was either a force or a couple. If the
resultant is a force, R, its action Hne is parallel to the action lines
of the forces of the system and its magnitude and sense are indi-
cated by the algebraic sum of the forces, that is, R = i:F. In
order to locate the hne of
action of the resultant force
the principle of moments will
be applied. For convenience
the origin, 0, will be taken
as the center of moments.
The moment of R then with
respect to 0 is equal to the
algebraic sum of the mo-
ments of the forces with
respect to the same point.
If the distance from the
action line of R to the 2/-axis is denoted by x, the moment of R
with respect to the origin is equal to Rx. Further, if the algebraic
sum of the moments of the forces of the system with respect to
the origin is denoted by 2(Fx) or Silf o the principle of moments
may be expressed by the equation Rx = 2iif o- The resultant, then,
if a force, is parallel to the 2/-axis and is determined by the following
equations,
R = ^F,
Rx=i:mo.
Y
i
1
1
i
Fe
Y''
U
'
/
'2
Fig. 37.
It should be noted that the sign of x is not always indicated prop-
erly by the ratio of XMq to R. The sign of x may be determined
by inspection, since the resultant force must he on that side of the
moment-center which will make the sense of its moment agree
with that of the moment of the system.
If the resultant of all, except one, of the forces of the system is
equal to that one and of opposite sense, XF then equals zero and
the resultant is a couple, the moment of which, according to the
ALGEBRAIC METHOD
41
principle of moments, is equal to the algebraic sum of the moments
of the forces of the system. Hence, if the resultant of a coplanar
parallel force system is a couple the moment, C, of the couple may
be determined from the expression,
the aspect of the couple, of course, being the same as that of the
plane of the forces.
If both SF and XM equal zero the resultant is equal to zero.
ILLUSTRATIVE PROBLEMS
26. Fig. 38 represents a beam resting on two supports and carrying four
loads as indicated. Find the resultant of the loads.
2000 1b.
1500 lb.
lAOOlb.
1000 lb.
-2-^
p-t-« 4 >i 1
•x=7.54-
FiG. 38.
Solution— R^-LF^- 1000 - 1500 - 1000 - 2000 = - 5500 lb.
In order to locate the action line of the resultant, the point A will be
selected as the moment-center. Thus,
RX = SMa
Therefore,
1000X2-1500X5-1000X8-2000X12= -41,500 lb .-ft.
- 41500 _ _ . ,,
X= =7.54 ft.
5500
Hence a single- load of 5500 lb. at a point 7.54 ft. from A will produce the
same external effect (reactions at A and B) as the four given forces,
42
RESULTANTS OF FORCE SYSTEMS
27. An arm mounted on an axle at 0 (Fig. 39) is acted on by the four forces
shown. Determine the resultant of the forces.
15 1b.
T
24 1b.
151b.
20 1b.
h
1
24 1b.
■< — h
101b.
Solution. — ■
i2 = 2F = 15 + 15 -20- 10=0.
Therefore the resultant is not a
force, but is a couple, C, the moment
of which may be found by taking mo-
ments about 0. Thus,
C = SMo = 20X10-15X6-10X5
+15X12 = +240 lb .-in.
Hence any couple having a moment
of +240 lb .-in. will produce the same
external effect on the arm (change in
the rotational motion) as the four given
forces. One resultant couple having forces equal to 24 lb. is shown in the
figure.
Fig. 39.
PROBLEMS
28. Fig. 40 represents a beam 12 ft. long acted upon by five forces as shown .
Find the anti-resultant of the five forces.
29. Five forces act on a body
2001b.
1501b.
501b.
A?
m7y
^100 lb.
as shown in Fig. 41. Replace
the two 60-lb. forces by a single
force and then find the resultant
of this force and the remaining
three forces.
30. A force of 10 lb. acts along
the 2/-axis, the sense of the force
being positive. Resolve the force
into components P and Q along
the lines x = l and a; = 3, respect-
ively.
Ans. P = 151b. 0=-5 1b.
31. A bar 6 ft. long is hinged
at A and is acted on by five forces as shown in Fig. 42
resultant of the five forces.
Ans. C= +90 lb. -ft.
32. A bar 10 ft. long is acted on by the forces shown in Fig. 43. Find
the equilibrant of the force system and represent it acting on the bar.
Ans. Equilibrant = +15 lb., 2f ft. to the left of the 20-lb. force.
33. The 40-lb. force shown in Fig. 44 is the resultant of the 10-lb. force and
a force P not shown. Determine P completely by means of a force and
a funicular polygon and check the result by the algebraic method.
801b.
Fig. 40.
mm
Determine the
ALGEBRAIC METHOD
43
In each of the following three problems find the resultant of the-iorce
system. The forces in each problem are parallel to the i/-axis. The values
80 1b. ^
rr —
\
\
/
/
\
\
/
/
/
/
\
\
\
^ 50 lb.
100 lb.
\
\
Fig. 41.
Fig. 42.
Fig. 43.
10 lb.
t
40 lb.
Fig. 44.
of F given are the magnitudes of the forces (expressed in pounds) and the
values of x are the distances of the action lines of the forces from the i/-axis
(expressed in feet).
34. F
+10
-20
+25
-15
X
- 2
+ 1
+ 3
+ 4
Ans. C=-25 Ib.-ft.
36. F
-10
+ 5
-20
+15
X
- 2
+ 1
+3
An&
+ 5
. 22= -10 lb. x = -4ft.
36. F
+40 .
-60
+80
-70 +10
X
0
+ 4
+ 9
+12 +14
44
RESULTANTS OF FORCE SYSTEMS
27. Couples. — In Art. 15 a couple is defined as two equal,
parallel, non-collinear forces of opposite sense. The moment of a
couple is defined as the algebraic sum of the moments of the two
forces of the couple, and it is shown that the moment is the same
for aU moment-centers in the plane of the couple. Further, it is
shown that the characteristics of a couple (moment, aspect, and
sense) may be completely represented by means of a vector.
Since couples play an important part in engineering problems it
will be of advantage to consider certain additional propositions
and transformations by means of which couples can easily be
combined with other couples or with other forces.
Proposition I. — A couple may be translated into any parallel
position in its oion plane or into any parallel plane without changing
its &xternal effect on the body on which it acts.
Proof. — Consider the couple Pp having a moment-arm, AB,
equal to p as shown in Fig. 45. Construct a parallelogram having
AB s,s one side and CD as
the opposite side. This par-
allelogram may or may not
be in the plane of the given
couple. At each of the
B points C and D introduce
two forces of opposite sense
which are parallel to the
<\ forces of the given couple and
^ ^P equal to them in magnitude.
The force P at A and the
upward force P at D may be
replaced by their resultant.
This resultant is an upward
force, 2P, the action line of
which is parallel to the ac-
tion lines of the forces of
the given couple and is locat-
ed half way between them.
Hence the action fine passes through 0, the point of intersection
of the diagonals of the parallelogram. Similarly, the force P at B
and the downward force P at C may be replaced by a downward
force, 2P, acting through 0 parallel to the forces of the given
couple. The two forces 2P, 2P at 0 cancel, thereby leaving only
COUPLES
45
the upward force P at C and the downward force P at Z). Thej»?ft
two forces constitute a couple which is equivalent to the given
couple. Hence the proposition is established.
Proposition II. — A couple may he rotated through any angle in
its plane without changing its external effect on the body on which ii
acts.
Proof. — In Fig. 46 the tw^o equal forces, Pi and P2, consti-
tute a couple having a moment-arm, AB, equal to p. Through 0,
the midpoint of the arm, draw any Une CD making an angle 6 with
AB and scale off OD and OC
each equal to ^p. At C in-
troduce two opposite forces,
P3 and P4, which are perpen-
dicular to CD, each of which
is equal in magnitude to the
forces Pi and P2 of the given
couple. Similarly at D intro-
duce two forces, P5 and Pe,
perpendicular to CD, each
being equal in magnitude to
the forces of the given couple.
Prolong the action Une of Pi
until it intersects the action
line of P4 at E. The resultant
of the two forces Pi and P4
is a force the action line of which bisects the angle AEC and hence
coincides mth EO. Let R denote the magnitude of this force.
In the same way the resultant of P5 and P2 may be shown to be a
force of magnitude R the action line of which is FO. Since these
two forces R, R are equal, colhnear, and opposite they cancel,
thereby leaving the forces P3 and Pe which constitute a couple
having a moment -arm CD which, by construction, is equal to p.
This equivalent couple, therefore, is merely the given couple
rotated through an angle d.
Proposition III. — The magnitudes of the forces of a couple and
the moment-arm of the couple may he changed to any values without
changing the external effect of the couple, provided that the product
of either force and the moment-arm remains constant.
Proof. — Given the couple Pp (Fig. 47) having a moment-arm,
AB, equal to p. Let the arm be increased to CD = {p-\-2x) as
Fig. 46.
46
RESULTANTS OF FORCE SYSTEMS
shown. At C introduce two forces of opposite sense each equal to
Pv
_/ and perpendicular to CD. Similarly, introduce two forces
Pv Pv
^ at Z). The force P at A and the upward force — r^ at jD
p+2x " ' -- - - - IT- p-\-2x
may be replaced by their resultant, which is an upward force of
2P(v-\-x)
magnitude — -jy: — , the direction of the line of action being per-
pendicular to CD. The position of the line of action of this result-
2P(p + x)
p -i-Zx
1
2x
1
.P
1
P +
^
\
X
%
p
%P B
0-
D
/
\
^
Pp
Pp
p + 2a;
>
P
p+ 2x
2P(j9+a;)
p + 2x
Fig. 47.
ant is found, by applying the principle of moments, to pass through
the mid-point oi AB. In a similar manner the resultant of the
force P at B and the downward force
p+2x
at C is found to be a
2P(v-\-x)
downward force of magnitude — ^^^ — - perpendicular to CD, the
action line of which also passes through the mid-point of ^P.
2P(v-\-x)
The two forces of magnitude — ~-^ — ~ balance, thereby leaving two
p-f-Zx
Pv Pv
forces, /I at C and at D, which constitute a couple the
p -7- Zx p -\- Zx
moment of which is equal to Pp.
Proposition IV. — The resultant of any number of couples in a
plane is a couple the moment of which is equal to the algebraic sum of
the moments of the couples.
COUPLES
47
This proposition can be proved by means of the three traas- —
formations discussed above. It is sufficient to prove the propo-
sition for two couples, for, if two couples can be combined into a
single resultant couple, evidently this couple can be combined with
a third coHiple, and so on. Let the two couples Pp and Qq (Fig.
48a) act on a body. The magnitude of the forces of the second
couple may be changed from Q to P by changing the moment-
arm froto q to -^ (Fig. 486), as discussed in Proposition III.
The second couple can then be rotated until the forces of the two
couples are parallel (Fig.
48c) . The second couple
can now be translated
until one of its forces is
collinear with one of the
forces of the first couple
(Fig. 4Sd). This trans-
lation can always be
made in such a way that
these two equal, collinear
forces are opposite in
sense and hence will can-
cel, thereby leaving the
two equal, parallel, and
opposite forces P, P which
constitute a couple. The
arm of this resultant
cou
Pleis(p+|)
moment is P(p+-
Pi
p
QQ
p^
, p
p
r
p . and its
Qq\ I ^ t ""P I (d)
. ~P)-
Pp + Qq. Hence the mo-
ment of the resultant is
equal to the sum of
the moments of the
given couples as stated in the proposition.
Proposition V. — Two coplanar couples which have the same
moment and sense are equivalent.
This proposition follows from the three transformations pre-
FiG. 48.
48 RESULTANTS OF FORCE SYSTEMS
viously discussed, since, by means of them, one of the couples
may be superimposed on the other.
§ 4. NON-CONCURRENTj NON-PARALLEL FoRCES IN A PlANE
28. Graphical Methods. First Method. — The resultant of a
system of non-concurrent, non-parallel forces in a plane is either
a force or a couple. If the resultant is a force it may be
determined by use of the parallelogram law. In Fig. 49 are
shown three non-concurrent, non-parallel forces Fi, F2, and Fz
acting on a body. The forces F\ and F2 may be combined into a
single resultant force Ri. R\ and Fz may be combined into a
Fig. 49. '
single force R2 which is the resultant of the given system of
forces.
If the resultant obtained by combining all, except one, of
the forces of such a system is equal to that one and is parallel to
it and of opposite sense, the two forces constitute a couple.
Furthermore, if the action lines of these two forces are colhnear
the forces cancel and the resultant of the system is equal to
zero.
Second Method. — The resultant of a sytem of non-concurrent,
non-paraUel forces in a plane may also be found by the con-
struction of a force and a funicular polygon. This method is of
great importance in engineering problems. The method of con-
struction of these polygons is the same as that described in con-
THE PRINCIPLE OF MOMENTS
49
nection with a system of parallel forces (Art. 24) and will not here-
be discussed in detail for the non-parallel force system. The
method will be illustrated, however, by the following problem.
ILLUSTRATIVE PROBLEM
37. A beam 9 ft. long is acted on by four forces as shown in Fig. 50.
Determine the resultant of the forces by use of the force and funicular
polygons.
Solution. — The force polygon as shown in Fig. 50(6) is constructed by
laying off the vectors AB, BC, CD, and DE which represent the magni-
tudes and the directions of the given forces. The closing side AE oi
the polygon represents the resultant force in magnitude and in direction.
By measuring ^^ to scale the magnitude of the resultant force is found
Scale. 1 in. = 8 ft.
(a)
Scale. 1 in. = 2000 lb.
(6)
Fig. 50.
to be 2450 lb. and the line AE is found to make an angle of 80° with the
horizontal.
In Fig. 50 (a) is shown the funicular polygon in which one point on the
action line of the resultant is found, namely, the intersection of oa and oe.
Therefore the action line of the resultant passes through this point and is
parallel to AE. By measurement the action line is found to intersect the
beam at a point 3.6 ft. from the left end of the beam. Hence, if the
four forces were replaced by a single force, R, of 2450 lb. as shown in Fig.
50(a), the reactions at the ends of the beam (external effects) would be
unchanged.
29. Principle of Moments. — The algebraic sum of the moments
of the forces of a coplanar, non-concurrent, non-parallel force
system about any point in the plane of the forces is equal to the
50
RESULTANTS OF FORCE SYSTEMS
moment of the resultant of the system about the same point.
The proof is identical with that given for a system of parallel
forces (Art. 25) and the principle will be used, in the algebraic
method, for finding the action line of the resultant force or the
moment of the resultant couple similar to the way it was used for
parallel forces (Probs. 26 and 27).
30. Algebraic Method. —The resultant of a system of non-con-
current, non-parallel forces in a plane is either a force or a couple
as explained in Art. 28. If the resultant is a force it may be deter-
mined as follows: In Fig. 51(a) is represented a body acted on by
four forces F\ F" , F'", and F"", Let each force be replaced
(6)
Fig. 51.
by its X- and ^/-components. The original force system is now
resolved into two systems of parallel forces, the resultant of
each parallel system being a force parallel to the forces of
the system. According to Art. 26 the resultant of the com-
ponents in the a;-direction is SFx and similarly, the resultant
of the components in the ^/-direction is l^Fy. Hence, the
magnitude and the direction' of the resultant of these two forces
(which is also the resultant of the original forces) may be
found from the equations R = V {1,Fxy-{-{1.Fyf and tan^^=|§^
(Fig. 516), in which dx is the angle which the action line of R
makes with the a;-axis. In order to obtain the position of the line
of action of the resultant force R the principle of moments may be
ALGEBRAIC METHOD 51
used. The principle is expressed by the equation Ra=XMjj^
in which a (Fig. 51a) is the perpendicular distance from the mo-
ment-center, 0, to the action Hne of the resultant, and 2lfo is
the algebraic sum of the moments of the forces with respect
toO.
Hence, if the resultant of a coplanar, non-concurrent, non-
parallel system of forces is a force, it may be determined completely
by the equations.
tan e^
^F,
If both XFx and XFy are equal to zero the resultant is not a
force and hence is a couple the moment, C, of which, according to
the principle of moments, is the algebraic sum of the moments of
the forces of the system, that is,
C = SM.
The center about which the moments of the forces are taken may
be any point in the plane of the forces since the moment of a couple
is the same about all points in the plane. The sense of rotation
of the resultant couple is indicated by the sign of the algebraic
summation and the aspect of the couple, of course, is the same as
that of the plane of the forces.
If ZFx and XFy are equal to zero and ZM is also equal to zero,
the resultant is equal to zero.
52
RESULTANTS OF FORCE SYSTEMS
ILLUSTRATIVE PROBLEM
38. Find the resultant of the system of four forces which act on the body
represented in Fig. 52(a). Each space represents 1 ft.
^
d
-/
II
/
/- .
2 F= 16.47 lb.
Solution. — The solution may be put in tabular form as follows:
F dx Fx = F COS 0x
20 45° 14.14
25 135° -17.67
5 270° 0.00
20 0° 20.00
Therefore,
Fy=F sin d,
14.14
17.67
- 5.00
0.00
Mo
28.28
0.00
-15.00
20.00
SFa; = 16.47 lb. SFy = 26.81 lb. 2Mo = 33.28 Ib.-ft.
R = V(16.47)2+ (26.81)2 = 31 .45 lb.,
0x=tan-i?^ = 58°25',
-3Ti5 = l-«^^^-
PROBLEMS
39. Fig. 53 represents a board 3 ft. square acted upon by four forces as
shown. Find completely the resultant of the forces. Specify the action line
of the resultant by the perpendicular distance from the origin 0.
Ans. ie = 47.6 lb. 0a; = 78° 10'. a = 1.98ft.
GRAPHICAL METHOD
53
40. Determine completely the resultant of the five forces acting on the—
body shown in Fig. 54. Each space represents 1 ft.
Fig. 53.
Fig. 54.
41. The forces of a coplanar system are specified below, the magnitudes of
the forces being expressed in pounds and distances x and y in feet.
F
^y y
Find the resultant of the system.
Am. 72 = 37.1 lb. 0, = 57°2O' a = 1.71 ft.
20
15
10
15
0, 2
0,2
4,2
6.2
0°
45°
90°
135°
§ 5. Concurrent Forces in Space
31. Graphical Method. — The resultant of a system of non-
coplanar, concurrent forces is a force which may be found by any
of the graphical methods used in finding the resultant of a system
of coplanar, concurrent forces as discussed in Art. 22. In applying
any one of these methods, the forces are projected on two of the
coordinate planes and the resultants of the two projected systems
of forces are found. The resultant of the original force system will
be a force the projections of which on the two planes are the
resultant forces of the projected systems. If the force polygon
54
RESULTANTS OF FORCE SYSTEMS
for each of the projected systems closes the resultant of the given
system vanishes. A graphical method of determining the resultant
of any non-coplanar system of forces is less convenient than is
the algebraic method of solution.
32. Algebraic Method. — Before discussing the method of
determining the resultant of any number of non-coplanar, con-
current forces the special
case of three concurrent
forces having action lines
which are mutually perpen-
dicular will be considered.
In Fig. 55 are represented
three such forces, P, Q,
and *S, the action lines of
the forces being taken as
the coordinate axes. The
forces P and Q may be
combined into a single force
OA the magnitude of which
is VW+Q^. The result-
ant of this force and the
force S is also the resultant of the given system of three forces
and is represented by the vector OB, its magnitude being
Vp2-|-Q2_|_^2 Hence the magnitude of the resultant, R, of the
three forces and the angles dxy dp, and dz, which the line of action
of the resultant makes with the coordinate axes may be found
from the equations,
Fig. 55.
r = VP^-\-Q^-{-S^,
cos dx = J^, COS dy
Q
R'
COS f^z = ^-
In finding the resultant of any number of non-coplanar, con-
current forces by the algebraic method it will be convenient to
take the point of concurrence of the forces as the origin of a set of
rectangular axes. Each force of the system may be resolved into
components along the coordinate axes (Art. 12). The system is
thus replaced by three collinear systems each of which may be
replaced by a single force (Art. 21). Thus, the resultant of the
components along the a:-axis is a single force along the a;-axis, the
ALGEBRAIC METHOD
55
magnitude of which is expressed by llFx. Similarly, the ^/-com:^
ponents may be replaced by a single force of magnitude ^Fy along
the i/-axis, etc. (Fig. 56). These three forces may be combined
into a single force which is the resultant of the given system and
which is completely defined by the following equations:
cos 0o.=
/? = V(SF,)2+(SF,)2+(SF,)2,
z
008 6^, = -^,
COS 62 = —^.
R
where dx, By, and Bz are the
angles which the action Hne
of the resultant makes with
the coordinate axes as shown
in Fig. 56.
Fig. 56.
ILLUSTRATIVE PROBLEM
42. Find the resultant of the following system of forces which pass through
the origin. The values of F are the magnitudes of the forces and the values
of X, y, z are the coordinates of points on the action lines of the forces.
F 50 lb. 90 lb. 100 lb. 60 lb. 20 lb.
X, y, z 5, 5, 0 7, 5, 3 3, 0, 4 2, 5, 4 0, 5, 3
Solution:
SFa; = 50-^+90.-^+100-?-+60--|=+20--^
V50 V83 5 V45 V34
= 35.4+69.2+60+17.9+0
= 182.5 lb.
SF:, = 50— ^+90--^+100-^+60-
V50 V83 5
= 35.4+49.3+0+44.7+17.1
= 146.5 lb.
-|=+20.^
V45 V34
SF. = 50.-^+90.-^+10o4+60-^+20.-4=
V50 V83 5 V45 V34
=0+29.6+80+35.8+10.3
= 155.7 lb.
56 RESULTANTS OF FORCE SYSTEMS
i2 = V(182.5)2 + (146.5)2+(155.7)2 = 281 lb.
182 5
0x = cos - 1 ^gp = cos - 1 . 650 = 49° 30'.
14fi *!
By = COS - 1 -^Ip = cos - 1 .521 = 58° 35'.
1^^ 7
0? = cos-i _^_ = cos-i .553=56° 25'.
PROBLEMS
In the following problems the forces are concurrent at the origin. It is
-equired to find the resultants of the systems.
43. F 10 lb. 20 lb. 15 lb.
X, y, z 1, 2, 1 2, 2, 3 3, 1, 2
Ans. i2 = 43.11b. 0^ = 53° 10'. 02, = 59°3O'. 0^ = 51° 50'.
44. F 100 lb. 150 lb. 50 lb. 200 lb.
X, 2/, z 2, 2, 1 3, 2, -2 -4, -3, -5 3,-2, 4
Ans. R = 27^\h. 0^ = 18° 10'. 02^=80° 45' 02 = 74° 20'.
46. F 10 lb. 30 lb. 20 lb.
X, y, z 1, 2, 1 1, 4, 2 2, 3, 3
§ 6. Parallel Forces in Space
33. Graphical Method. — The resultant of a system of non-
coplanar, parallel forces (see Fig. 57) is either a force or a couple.
If the resultant is a force the action hne is parallel to the forces
of the system, its magnitude and sense being given by the algebraic
sum of the forces. The resultant force may be found graphically
by a repeated use of the first method described in Art. 24 for finding
the resultant of two parallel forces. Since this method of solu-
tion, however, involves constructions in various planes it is not
convenient to use. The resultant force may also be found by the
construction of a force and a funicular polygon as described in
Art. 24. In using this method the forces are projected on two of
the coordinate planes and the resultant of each of the two systems
of projected forces is found. The resultant of the given force
system is a force the projections of which on the two coordinate
planes are the two resultants of the projected forces.
ALGEBRAIC METHOD
57
34. Principle of Moments. — The algebraic sum of the moments
of any nmnber of non-coplanar, parallel forces about any line in
space is equal to the moment of their resultant about the same
line. The proof of this proposition is similar to the proof of the
principle of moments given for a coplanar, parallel force system
and will be left to the student. This principle will be used in the
algebraic method of determining the resultant of a non-coplanar,
parallel system in much the same way that it was used in Arts.
26 and 30.
35. Algebraic Method. — In determining the resultant of a
system of non-coplanar, parallel forces by the algebraic method it
is convenient to select coordinate axes so that one axis is parallel
to the forces. In Fig. 57 is shown a sj^stem of parallel forces
referred to such a set of axes. The resultant, if a force, is parallel
to the forces, its magnitude, R, being equal to the algebraic sum of
the forces. The line of action of the resultant force is found by
applying the principle of moments. Thus, if the algebraic sum
of the moments of the forces with respect to the a;-axis be denoted
by 2Mx and the distance of the resultant from the x-axis be
denoted by y, then the principle of moments is expressed by the
equation Ry = ^Mx. In a similar manner, Rx='EMy. Th^^
resultant, if a force, will then be completely defined by the fol-
lowing equations:
R = ^F,
2My
R '
R '
If the resultant of all,
except one, of the forces
of a parallel system is a
force which is equal to that
one but of opposite sense,
then XF equals zero and hence the resultant is not a force and is
therefore a couple. According to the principle of moments the
moment, Cx, of the resultant couple with respect to the x-axis is
58
RESULTANTS OF FORCE SYSTEMS
equal to the algebraic sum of the moments of the forces with
respect to the a:-axis, that is,
Similarly,
Cy=i:My.
Thus, the resultant is a couple which Hes in a plane parallel to the
2-axis, the moments of which, with respect to the x- and 2/-axes, are
Cz and Cy as expressed above. Methods of expressing a resultant
couple in terms of its component couples are discussed in Art. 36.
ILLUSTRATIVE PROBLEM
46. Find the resultant of the system of parallel forces as shown in Fig. 58.
Each space in the figure represents 1 ft.
Solution :
R»25 1b. I ^30 lb.
Fig. 58.
72 = 2^ = 10 + 15-20-30= -25 lb.
SM:r = 20X2+30X3 -10X1 -15X1.
= 105 lb .-ft.
SiWy = 10X1+15X3 -20X2-30X5.
= -1351b.-ft.
ALGEBRAIC METHOD
59
and
5 = f=5fft.
- 105 ,, ,^
Hence the resultant is a downward force of 25 lb. as shown in Fig. 58.
Caution. — Care must be exercised in finding x and y. For instance, in the
above example if the value of SMj; ( + 105 lb .-ft.) be divided by R{ — 25 lb.)
the quotient is — 4i ft., which is not the value of y, since a downward force of
25 lb. in this position would have a moment of — 105 lb .-ft. with respect to the
a;-axis. The magnitudes of x and y should be obtained by dividing the mag-
nitudes of XMy and SMa, by the magnitude of R and their signs should be
determined by inspection. The signs of x and y, of course, must be such that
the moment of the resultant will have the same sense of rotation as indicated
by the algebraic sum of the moments of the forces of the system,
PROBLEMS
47. A board 6 ft. square is acted on by five forces as shown in Fig. 59.
Determine the resultant of the forces.
Ans. R = +10 \h. ^ = +2.5ft. y = +3.5ft.
60 1b.
Fig. 60.
48. A table 5 ft. square carries four concentrated loads as shown in Fig. 60.
Find the resultant of the four loads.
Ans. R = lS5\h. x = +2.04 ft. y = +2.26 ft.
49. Find the resultant of the following system of forces which are parallel
60
RESULTANTS OF FORCE SYSTEMS
to the z-axis. The values of x and y, expressed in feet, are the coordinates
of the points where the action lines of the forces intersect the a:?/-plane.
F 20 lb. 10 lb. 25 lb. -15 lb. -10 lb.
X, y
1, 3
3. 2
6, 6
2, 5
6. 4
§ 7. Couples in Space
Proposition, — The
36. Resultant of a System of Couples.
resultant of any number of couyles is a couple.
Proof. — It is sufficient to prove
this proposition for two couples,
only, since if two couples can be com-
bined into a single resultant couple
this couple can be combined with a
third couple in exactly the same way
and so on. Thus, consider the two
couples Pp and Qq in planes making
an angle a with each other as shown
in Fig. 61(a). The forces Q, Q of
the couple Qq can be made equal to
P, P if the arm be changed to -^
(Art. 27) as shown in Fig. 61(6).
Each couple can then be rotated
in its plane until the forces of the
couples are parallel to the line of
intersection of the two planes (Art.
27) as shown in Fig. 61(c). Now let
the two couples be translated until
one force of each couple lies in the
line of intersection of the two planes.
This translation can always be made
so that the two forces in this hne
are opposite in sense and hence will
cancel, thereby leaving a couple the
forces of which are P, P (Fig. 6 Id).
The arm of this couple from trigo-
nometry is,
Fig. 61.
yjp
^+51-2^^ cos «.
Special Cases.— 1. If the angle a equals 90°, that is, if the
RESULTANT OF A SYSTEM OF COUPLES
61
planes of the couples are perpendicular (Fig. 62), the moment, C, .
of the resultant couple is
C = P^p'+%<f = V(Pp)2+(Qg)2.
That is, the moment of the resultant couple is the square root of
the sum of the squares of the moments of the two couples. The
plane of the resultant couple makes
an angle 4> (Fig. 62) with the plane of
the couple Pp such that
Q
tan0 = -- = ^-,
and the sense of rotation of the re-
sultant couple is indicated in Fig. 62.
II. If a equals 180°, the couples
are in the same or in parallel planes
and hence have the same aspect.
The moment of the resultant couple
then is,
C = P^lv'+%q'+2vlq = p{v-^qfj-Pv+Qq•
That is, the moment of the resultant couple is the algebraic sum
of the moments of the two couples and its aspect, of course, is the
same as that of each of the couples, the sense of rotation being
indicated by the sign of the algebraic summation.
Fig. 62.
Fig. 63.
Resultant of a Couple and a Force. — A couple and a force not in
the plane of the couple may be replaced by two non-coplanar
forces. Thus, consider a couple Pp in the plane MN and a force F
62
RESULTANTS OF FORCE SYSTEMS
as shown in Fig. 63(a). The couple may be translated according
to Art. 27 until one of the forces of the couple Pp and the force F
are concurrent (Fig. 636) . Therefore P and F now may be replaced
by their resultant R and hence the given system is replaced by a
force P in the plane of the original couple and a force R not in the
plane. Conversely, any two forces not in the same plane may be
replaced by a force and a couple.
37. Composition of Couples by Means of Vectors. Propost-
Hon. — // any number of couples he represented by their vectors, the
resultant of these vectors mil represent completely the resultant
couple.
It will be sufficient to prove the proposition for two couples.
The extension of the proof to any number of couples is obvious.
Proof. — Given two couples Pipi and P2P2 in planes which
make an angle a with each other as shown in Fig. 64(a). It is
y
/ J^/
/"
/
/
ia)
Fig. 64.
assumed that the couples have been transformed (see Art. 36)
so that P\ equals P2. Let DOE (Fig. 646) represent a cross-section
of the two planes shown in Fig. ^^.{0). The couple P\p\ can be
represented by a vector OA perpendicular to the plane of the
couple, the length of OA being proportional to P\p\. Likewise,
the couple P2P2 can be represented by the vector OB perpendicular
to the plane of the couple, OB being proportional to P2P2- Con-
struct the parallelogram OACB. In order to show that OC repre-
sents completely the resultant couple it must be shown that OC is
perpendicular to the plane of the resultant couple (which is the
same as showing that it is perpendicular to DE) and also that its
COMPOSITION OF COUPLES BY MEANS OF VECTORS 63
length represents to scale the moment, Pj), of the resultant couple;
In the two triangles OAC and ODE the angle OAC equals the angle
DOE. The vector OA is proportional to P\p\. This fact is
expressed by the equation,
Similarly,
Hence,
OA=kPm,
0B = kP2P2 = AC.
OA kPipi pi OD
AC kP2P2 P2 OE'
Therefore, the triangles are similar and hence OC is proportional to
DE. That is,
OC OA kPpi
DE OD vi
Therefore,
OC = DEXkP = kPp.
Hence OC represents the moment of the resultant couple. Further-
more, OC is perpendicular to DE, for, since the triangles OAC and
ODE are similar and the corresponding sides OA and OD are per-
pendicular, it follows that the corresponding sides OC and DE are
perpendicular. Hence, OC represents the aspect as well as the
magnitude of the resultant couple. It is plain from Fig. 64(6)
that OC also represents the sense of the resultant couple. Any
number of couples then may be combined into a resultant couple by
representing each couple by a vector and finding the resultant of
the system of vectors. The resultant vector will represent com-
pletely the resultant couple.
For an application of this method of combining couples see
the discussion of the problem of balancing rotating masses, such as
an engine crankshaft, in Chapter IX, Section III.
Three Couples in Mutually Perpendicular Planes. — Let the
given planes be taken as the coordinate planes. The couple lying
in the yz-plsme may be represented completely by a vector along
the a;-axis. Let this vector be denoted by Cx. Similarly, the
couples in the xz- and x^z-planes may be represented by vectors
Cy and Cz along the y- and 0-axes respectively. The resultant of
the three couples will be represented by the resultant of the three
64
RESULTANTS OF FORCE SYSTEMS
vectors Cx, Cy, Cg. If the resultant of these three vectors be
denoted by C the resultant may be found from the equation,
Also, if </>!, </)y, and <j5>z be the angles which the vector C makes with
the coordinate axes, these angles may be found from the equations,
C
0y = cos-i-^,
<f>z = COS~^—.
That is, if the couples Cx, Cy, and Cz, which act on a given body,
are replaced by the couple C acting in the plane defined by the
angles c^i, <l>y, and <^2, the external effect on the body will be
unchanged.
ILLUSTRATIVE PROBLEM
50. Determine the resultant of the three couples which act on the body
(cube) as shown in Fig. 65. Each of the edges of the cube is 4 ft. in length.
Solution. —
Cx = 200X4 = 800 lb. -ft.
Cy = 300X4 = 1200 Ib.-ft.
^2 = 100X4= 4001b. ft.
Therefore,
C = ^(800) 2+ (1200) 2+ (400) 2.
= 1500 lb. -ft.
. 800
Z
300 lb.
y
/ 100 lb.
300 lb.
X
/
/^
/ 100 lb.
COS-
COS"
/v
Fig. 65.
</)z = COS~^
1500
1200
1500
400
1500
= 57° 47',
= 36° 50',
= 74° 30'.
Hence a single couple having a moment of 1500 lb .-ft. and located in a
plane perpendicular to the line defined by the above angles will have the same
external efifect on the bod^^ as the three given couples.
38. Resolution of Couples. — In order to resolve a couple into
two or more component couples, the vector representing the given
couple may be resolved into component vectors each of which
represents a component couple.
RESOLUTION OF COUPLES 65
One of the most important cases of the resolution of a couplers
that in which the couple is resolved into three component couples
lying in planes which are mutually perpendicular. Let the planes
of the component couples be taken as the coordinate planes and
let <f)x, (f>v, and 0z be the angles which the vector representing the
given couple, C, makes with the coordinate axes. If Cx, Cy, Cg, be
the component couples lying in the yz-, zx-, and xy-planes,
respectively, they may be found from the following equations:
Cx=(7cos0i, Cv = (7 cos <^v, Cg = C cos 4>z.
§ 8. Non-concurrent, Non-parallel Forces in Space
39. Graphical Method. — The resultant of a system of non-
concurrent, non-parallel forces in space may be a force and a couple,
a single force, or a single couple. Further, since a force and a
couple, not in the same plane, may be replaced by two non-
coplanar forces (Art. 36), the resultant of such a system may also be
regarded as two non-coplanar forces. If the graphical method is
used in determining the resultant of such a system of forces, it is
convenient to reduce the system to two forces rather than a force
and a couple. The resultant may be found by selecting arbitrarily
any plane which is not parallel to any of the action lines of the
given forces and resolving each force, at the point where its action
line pierces this plane, into two components, one perpendicular
to this plane and one lying in the plane. The system consisting
of the components which lie in the plane may be combined accord-
ing to the method of Art. 28 and the components which are per-
Vsndicular to the plane may be combined according to the method
of Art. 33. If the resultant of each of these systems is a force
these two forces may be regarded as the resultant of the given
system. The two forces may be replaced, however, by a force and
a couple. Further, in special cases the force may vanish and
thereby leave a single couple as the resultant of the given system.
Again, the couple may vanish, in which case the resultant of the
system is a force. If both the force and couple vanish the given
force system has no resultant.
40. Principle of Moments. — The algebraic sum of the moments
of the forces of any non-coplanar, non-concurrent, non-parallel
system about any line is equal to the moment of the resultant
66 RESULTANTS OF FORCE SYSTEMS
of the system about the same line. Since the resultant in general
consists of a force and a couple, the moment of the resultant
about any Une must be regarded as the algebraic sum of the
moments of the force and the couple about the Hne.
41. Algebraic Method. — When the resultant of a system of
non-concurrent, non-parallel forces in space is a force and a couple,
the force may be made to pass through any arbitrary point, but,
for different positions of this point, the moment of the couple will
vary. In determining the resultant force and couple by the alge-
braic method it is convenient to select a rectangular set of coor-
dinate axes so that the origin is the arbitrary point through which
the resultant force passes. Each force of the system may be
replaced by an equivalent parallel force through the origin and a
couple as in Art. 18. Thus the system is reduced to a system of
concurrent forces through the origin and a system of couples.
The concurrent system at the origin may be combined into a
resultant force, as in Art. 32, which is completely defined by the
following equations:
R = V(2F,)2+(2F,)2+(2F,)2,
(9x = cos-i -^,
^j, = cos-i-^,
^, = cos-i^.
The system of couples may be replaced by a single couple as in
Art. 37. For convenience in determining the resultant couple it
will bo considered to be resolved, as in Art. 38, into three com-
ponent couples lying in the coordinate planes. Let the couple
lying in the zy-plane be denoted by Cx, since it may be represented
by a vector along the x-axis. The vector which represents this
couple will also be denoted by Cx- Similarly, let the couples
in the other two planes, as well as the vectors which represent
them, be denoted by Cp and Cz. The given system then is equiva-
lent to the system shown in Fig. 66, namely, a force R through the
origin and the couples Cx, Cy, and Cz. Since the two systems are
equivalent, the sum of the moments of the forces of the given sys-
tem about any Une is equal to the sum of the moments of the
ALGEBRAIC METHOD
67
forces of the system shown in Fig. 66 about the same Kne. ThtiST^
let moments be taken about the .r-axis and let the moment of
the given system about the x-axis be denoted by XMx. The only
part of the system shown in Fig. 66 which has a moment about
the a:-axis is the couple lying in the ye-plane, and the moment of
this couple is Cx- Hence, i:Mx=Cx. Similarly i:My = Cy and
XMz = Cz. The resultant of the three couples then may be deter-
mined as in Art. 36. Thus the resultant couple is defined by the
following equations :
C = V(SM,)2+(SM,)24-(2Af,)2,
<f>x = COS
<f>P = COS
<t>z = COS
C '
C
_, Silf,
c
Fig. 66.
All of the force systems which have been discussed above
may be considered as special cases of this system, and the six
equations which define the resultant of this system will reduce to
the same equations which were found necessary to define the
resultants of the simpler systems.
ILLUSTRATIVE PROBLEM
51. Find the resultant of the system of four forces which act on the cube
as shown in Fig. 67. Each side of the cube is 4 ft. long.
Solution: In resolving the 10-lb. force into its three rectangular compo-
nents, A will be taken as the point of application of the force and at this
point the force may be resolved into two components, one along AC and one
along AB. The latter component may be resolved into components along
AE and AF. The quantities needed in the solution may be put in tabular
form as follows, forces being expressed in pounds and moments in pound-feet.
F
Fx
Fy
Fz.
Mx
My
M,
10
-4.67
-6.25
-6.25
0.00
—6.25
6.25
15
0.00
-10.60
-10.60
0.00
-42.42
42.42
18
18.00
0.00
0.00
0.00
-72.00
18.00
40
0.00
40.00
0.00
40.00
0.00
0.00
2;F;j = 13-33 2Fy =23.15 SFz = —16.85 SAfx =40.0 2My = —120.7 23/2=66.
68
RESULTANTS OF FORCE SYSTEMS
1
B (
r
/
^ •
1^ i
8 1b. /
/
\
\
1 \
1 \
f
0 ^
^>^6.
X
10 \h.\^
\
\
\
\
\
\
/
Fig. 67.
i2 = ^(13.33)24-(23.15)2+(16.85)2
= 31.6 lb.
13 33
0x=cos-i-^^=65°O'
31. o
23 15
0j,=cos-'^T^ = 42°5O'
* 31.6
02=008-'^^^ =122° 15'
C =^(40)2 + (120.7)2 + (66.66)'
= 144 lb. -ft.
40
0^ = cos-ij^ = 73°5O'
— 120 7
0» = cos-^ ,,:' = 146° 55'
</)2 = cos-
144
,66.66
144
= 62° 25'.
Note. — In specifying the angles which the vectors representing the result-
ant force and the resultant couple make with the coordinate axes, the smaller
of the two angles which each vector makes with the positive end of the given
axis is specified.
PROBLEMS
62. Find the resultant of the three forces which act on the cube as shown
in Fig. 68. Each side of the cube is 4 ft. long.
Fig. 68.
Fig. 69.
53. The sides of the cube shown in Fig. 69 are each 4 ft. long. Find the
resultant of the three forces which act on the cube.
CHAPTER III
EQUILIBRIUM OF FORCE SYSTEMS
§ 1. Introduction
42. Preliminary. — In the preceding chapter equations and
graphical constructions were found by the use of which the result-
ants of the various force systems may be determined. In the pres-
ent chapter are determined the algebraic and graphical conditions
which the forces of the various force systems must satisfy in order
that the resultants of the force systems shall be equal to zero;
that is, in order that the force systems shall be in equihbrium.
If a force system which is in equiHbriima acts on a body, the body
is either at rest or has a uniform motion.
The independent equations which must be satisfied by the
forces which hold a body in equilibrium are called the equations of
equilibrium, and the graphical constructions which the forces must
satisfy are sometimes called equilibrium diagrams or equilibrium
polygons.
Many problems in engineering practice involve bodies which
are in equilibrium under the action of a system of forces as, for
example, a bridge, roof -truss, crane, etc. In such problems there
may be certain elements of the forces acting on the body which are
unknown as, for example, the magnitude or the direction of one or
more of the forces. These unknown elements or quantities may
be found if their number is not greater than the number of equa-
tions of equiUbriima for the force system involved. Such force
systems are said to be statically determinate. If the number of
unknown quantities in a force system is greater than the nimiber
of equations of equilibrium for that particular force system, the
force system is said to be statically indeterminate, as, for example,
the forces which act on a horizontal beam which rests on three or
more supports and carries known vertical loads. The beam is in
equilibrium under the action of a system of coplanar, parallel forces
all of which are known except the three upward reactions of the
supports. As is shown in Art. 49, there are only two independent
70 EQUILIBRIUM OF FORCE SYSTEMS
equations of equilibrium for such a force system, and hence
the three reactions cannot be found from the equations of
equiUbrium. The force system is therefore statically indeter-
minate.
43. Graphical Conditions of Equilibrium. — In the previous
chapter it was shown that the resultant of an unbalanced force
system in a plane is either a force or a couple. Further, it was
shown that if the resultant is a force, it is represented in magni-
tude and in direction by the closing side of the force polygon, and
that if the resultant is a couple, the two forces of the couple act
along the first and last strings of the funicular polygon. Hence,
if the force polygon closes, the resultant cannot be a force, but
may be a couple. If, however, the funicular polygon also closes,
that is, if the first and last strings along which the two forces of
the couple act are colhnear, the two forces cancel and hence the
resultant couple vanishes. Hence there are two conditions which
the forces of a coplanar force system must satisfy if they have no
resultant, that is, if the forces are in equilibrium.
(1) The force polygon must close. If this condition is satis-
fied the resultant cannot be a force.
(2) The funicular polygon must close. If this condition is
satisfied the resultant cannot be a couple.
The conditions of equilibrium for non-coplanar force systems
may be stated in a similar manner. In order to determine the
resultant of a non-coplanar system graphically, the forces of the
system are projected on two of the coordinate planes and a force
and a funicular polygon is drawn for each of the projected systems.
The conditions of equilibrium for a non-coplanar system, then,
are that the force and funicular polygons for each of the projected
systems must close.
44. Algebraic Conditions of Equilibrium. — The two conditions
which the graphical diagrams for a balanced force system must
satisfy as stated in the preceding article may also be expressed
algebraically. Thus, if the force polygon closes, the projections
(components) of the forces on any fine also form a closed polygon,
as shown in Fig. 70, and since these components are collinear their
vector sum is the same as their algebraic sum. Hence, the fi^ct
that the force polygon for the components closes may be expressed
by stating that the algebraic sum of the components is equal to
zero. Therefore, if the force polygon for the given system of
ii
ALGEBRAIC CONDITIONS OF EQUILIBRIUM 71
forces closes, the algebraic sum of the components of the forces in
any direction is equal to zero.
If the funicular polygon closes, the resultant cannot be a couple,
since the first and last strings of the funicular polygon are collinear
and hence the algebraic
sum of the moments of ^ ^
the two equal and opposite
forces which act along
these strings is equal to
zero. But the algebraic
sum of the moments of
these two forces is equal to
the algebraic sum of the
moments of the forces of Yig. 70.
the system. Therefore,
the statement that the funicular polygon must close is equivalent
to the statement that the algebraic sum of the moments of
the forces of the system must equal zero. Hence the algebraic
conditions of equilibrium are :
(1) The algebraic sum of the components of the forces in any
direction must equal zero.
(2) The algebraic sum of the moments of the forces about any
axis must equal zero.
An infinite number of equations could be written in accordance
with these conditions by taking different directions of resolution
and different moment axes, but all of the equations would not be
independent. The number of independent equations is different
for the various force systems, as will be discussed in the succeeding
articles, but for any force system the independent equations of
equiUbrium are the equations which are necessary and sufficient
to ensure that the resultant for that particular force system shall
be equal to zero.
If a given body is in equilibrium under the influence of a
system of forces some of which are unknown, wholly or in part,
these unknown elements may be found by applying the equations
of equilibrium which apply to that particular system of forces.
If the number of unknown elements in a system of forces which
is in equiUbrium is. equal to the number of independent equations
of equilibrium for that particular system, the determination of all
of the unknown elements involves the use of all of the equations of
72 EQUILIBRIUM OF FORCE SYSTEMS
equilibrium. Frequently, however, it is not required to determine
all of the unknown elements in such a system, for, a single element
only may be required, as for example, the magnitude of a certain
force, the Une of action of which is known. In such cases the un-
known element may frequently be found by using only one of the
equations of equiUbrium. In applying the equilibrium equations
the work may be materially simplified by properly selecting the
directions of resolution and the axes of moments. Before applying
the equations of equilibrium to any system of forces which holds
a body in equilibrium it is important to have a clear idea of the
forces which act on the body. For this purpose a free-body dia-
gram is drawn. A free-body diagram is a diagram of a body
showing the actions of all other bodies (forces) on the body con-
sidered. It does not show the actions of the given body on other
bodies.
§ 2. CoLLiNEAR Forces
45. Equations of Equilibriimi. — A system of collinear forces
is in equiUbrium if the forces of the system satisfy either of the
following equations :
^F = 0, (1)
or
SM^ = 0, (2)
where A is any point not on the action line of the forces.
Proof. — As shown in Art. 21, if a colhnear force system is not
in equilibrium, the resultant of the force system is a force having
the same action line as the forces and having a magnitude, R, which
is given by the equation, R = 2F. If, then, the equation 'EF = 0,
is satisfied the resultant is not a force and therefore the system is
in equiUbrium. The equation 2ilf a = 0 is also sufficient to ensure
equiUbrium, for, in order to satisfy this equation, the resultant force
must pass through the point A . But this is impossible, since the
resultant force, if there be one, has the same line of action as the
forces and hence cannot pass through A .
Therefore, if either one of the equations (1) and (2) is satisfied,
the resultant is equal to zero and hence there is but one inde-
pendent equation of equiUbrium for a colUnear force system.
EQUATIONS OF EQUILIBRIUM 73
ILLUSTRATIVE PROBLEM ~
54. Two men pull on a rope with forces of 100 lb. each (Fig. 71o). What
is the stress in the rope?
Solution. — Suppose the rope to be divided into two parts A and B as
shown in Fig. 71(6). Consider as a free-body the part A. The forces acting
100 lb. , ! , 100 lb.
1 (^)
100 lb. ^ s s , B 100 lb.
(6)
Fig. 71.
on A are two in number, namely, the 100-lb. force and the force exerted by B
on A. The latter is the internal stress required. Let this stress be denoted
by S. The equation of equilibrium then becomes,
SF = aS-100 = 0.
Therefore,
^ = 100 lb.
Obviously, B could have been taken as the free body and the same result
would have been obtained.
§ 3. Concurrent Forces in a Plane
46. Equations of Equilibrium. — A system of coplanar, concur-
rent forces is in equilibrium if the forces of the system satisfy the
following equations:
(A)
where x and y denote any two non-parallel lines in the plane. It
is convenient, however, to take as the two Knes a set of rectangular
axes with the point of conciu-rence of the forces as origin.
Proof. — In Art. 23 it was shown that if a concurrent system of
forces in a plane is not in equiUbrium the resultant is a force, the
components of which are equal to llFx and XFy. If, then, the
forces of the system satisfy the equation SFx=0 the resultant
cannot have a component along the x-axis and if the equation
2i^y = 0 is satisfied the resultant cannot have a component along
the y-axis. Therefore if both of these equations are satisfied, the
resultant cannot be a force and hence the system must be in
74
EQUILIBRIUM OF FORCE SYSTEMS
equilibrium. There are, then, only two independent equations
of equilibrium for a coplanar, concurrent system of forces.
Another set of independent equations which, if satisfied by the
forces of a coplanar, concurrent force system, are sufficient to
ensure equilibrium may be expressed as follows :
(B)
where x denotes any line in the plane (taken for convenience as one
of two rectangular axes through the point of concurrence of the
forces) and A is any point in the plane not on the y-suds.
Proof. — If the equation ZFx = Ois satisfied, the resultant cannot
have a component along the a:-axis, that is, the resultant, if there
be one, must lie along the y-sods. If the equation ZMa = 0 is sat-
isfied, the resultant, if there be one, must pass through the point A.
It is impossible for a force to satisfy the two equations simultane-
ously, and hence if both of the equations are satisfied the system
is in equihbrium.
A third set of equations of equilibrium for a coplanar, con-
current force system is as follows :
(C)
where A and B are any two points in the plane of the forces, pro-
vided that the line joining A and B does not pass through the point
at which the forces are con-
current. The proof that
these equations are sufficient
to insure equihbrium will
be left to the student.
47. Lami's Theorem. —
When a coplanar, concur-
rent force system consists
of only three forces, the
equations of equihbrium
may be expressed in a spe-
cial form known as Lami's
theorem. Let Fig. 72(a) represent three concurrent forces in
equilibrium. The force polygon for the three forces is shown
Fig. 72.
THREE FORCES IN EQUILIBRIUM
75
in Fig. 72(6-). Since each side of a triangle is proportional t^ —
the sine of the opposite angle, the following equations are
obtained from the force polygon:
Fi
That is,
sin (tt— cci) sin (x— 0:2) sin (r—as)
Fi F2 F3
sin ai sin 0:2 sin 0:3
Hence, if three concurrent forces are in equihbrium, the magnitude
of each force is proportional to the sine of the angle included
between the action lines of the other two. This statement is
known as Lami's theorem.
y- 48. Three Forces in Equilibrium. — If three forces are in equihb-
rium the forces must be coplanar and either concurrent or parallel.
In order that the three forces shall be in equilibrium, the resultant
of any two of the forces must be a force which is collinear with the
third force, of equal magnitude, and of opposite sense. Now,
the resultant of the two forces will have the same line of action as
the third force only if the two intersect on the action hne of the
third force, or if the two forces are parallel to the third force.
Hence, the three forces must be either
concurrent or parallel. This principle
is of considerable importance, as it
simplifies the solution of many prob-
lems. Consider, for example, the
crane shown in Fig. 73(a). The forces
acting on the crane are the reaction
Ri at the upper end (assumed to be
horizontal), the load W, the weight of
the crane (not shown), and the reac-
tion R2 at the lower end, the direction
of the latter force being unknown.
The load W and the weight of the crane may be replaced by a
single resultant force R, and the system will then consist of three
forces Ri, R2, and R. Since the three forces must be concurrent,
R2 must pass through the point of intersection of Ri and R, and
hence its action line is determined as indicated by the dotted line.
The magnitudes of the reactions Ri and R2 may now be deter-
mined by drawing the force polygon (Fig. 736). The force polygon
Fig. 73.
v/
76
EQUILIBRIUM OF -FORCB SYSTEMS
is constructed by drawing AB to represent the known force R and
by drawing from A and B lines parallel to Ri and R2, respectively,
which intersect at C. The reaction Ri is represented by CA and
R2 is represented by BC.
ILLUSTRATIVE PROBLEMS
56. A body is held in equilibrium by a system of three concurrent forces
as shown in Fig. 74, Find the values of P and 6.
Fig. 74.
Solution. — The equations of equilibrium are,
'EFx = P cos 0-10 cos 30° -20 cos 45° =0.
/. P cos 0 = 8.66+14.14 = 22.80 lb
SFy = P sin 0+10 sin 30° -20 sin 45° =0.
.*. P sin 0 = 14.14-5.00 = 9.14 lb
By dividing (2) by (1), 0 may be obtained. Thus,
9.14
tan0:
.401.
(1)
(2)
22.80
.*. 0 = 21° 50'.
By squaring and adding (1) and (2) P may be obtained. Thus,
P2 = (22 .80) 2 +(9. 14) 2 = 603.5,
/. P = 24.5 lb.
THREE FORCES IN EQUILIBRIUM
77
56. Fig. 75(a) represents a lower panel point of a pin-connected PratL
truss. The stresses in two of the members are 1000 lb. and 3000 lb. as shown.
Find the stresses, P and Q, in the other members.
Algebraic Solution.
From (2),
Substituting in (1),
Fig. 75.
^Fx = Q+P cos 45° -3000 = 0,
(1)
(2)
P =
1000
1414 lb.
sin 45°
Q = 3000-1414cos45°.
= 3000-1000 = 2000 lb.
Graphical Solution. — The problem may be solved graphically by con-
structing a closed force polygon as shown in Fig. 75(6). The polygon is con-
structed as follows: Vectors AB and BC are drawn to represent the 3000-lb.
and 1000-lb. forces respectively. A line is then drawn from C parallel to the
direction of the force Q and a line is drawn from A parallel to the direction of
the force P. These lines intersect at D. Q is then represented by CD and
P by DA. The magnitudes of Q
and P are found by measuring,
according to the scale indicated,
to be 20001b. and 14101b., respect-
ively.
PROBLEMS
57. A sphere weighing 100 lb.
rests between two smooth planes
as indicated in Fig. 76. Find the
reactions of the planes on the
sphere.
Ans. Pi = 51.7 lb.; i22 = 73.1 lb.
Fig. 76.
78
EQUILIBRIUM OF FORCE SYSTEMS
58. A body weighing 80 lb. is supported by means of a number of cords
as shown in Fig. 77. Find the tensions in the cords and the vakie of d.
751b
80 lb.
Fig. 77.
Fig. 78.
59. Two bodies weighing 75 lb. and 100 lb. rest on a smooth cylinder and
are connected by a cord as shown in Fig. 78. Find the reactions of the cylin-
der on the bodies, the tension in the cord, and the value of 0.
§ 4. Parallel Forces in a Plane
49. Equations of Equilibriiun. — A coplanar, parallel force
system is in equilibrium if trie forces of the system satisfy the
equations,
{A)
where A is any point in the plane of the forces.
Proof. — According to Art. 26, the resultant of a coplanar,
parallel force system which is not in equiUbrium is either a force
or a couple. If the resultant is a force the magnitude, R^ is
expressed by the equation R = 'EF, and if the resultant is a
couple the moment, C, is expressed by the equation C = 'ZM.
If SF = 0 the resultant is not a force and if ZMa = 0 the
resultant is not a couple. Hence, if both equations are satisfied
the resultant of the force system can be neither a force nor a
couple and therefore the system is in equilibrium. Two equa-
tions, then, are necessary and sufficient to ensure that the forces
are in equilibrium. In other words, there are only two indepen-
EQUATIONS OF EQUILIBRIUM 79
dent equations of equilibrium for a system of parallel forces in ar-
plane.
Another set of equations of equilibrium for a system of coplanar,
parallel forces may be written as follows:
^^-M, (B)
where A and B are any two points in the plane, provided that the
line connecting A and B is not parallel to the forces of the system.
The proof that these equations are sufficient and necessary to
ensure that the forces are in equilibrium will be left to the student.
An important problem in the equilibrium of coplanar, parallel
forces is one in which the magnitudes of two forces are required,
all of the other elements of the forces being known. The graphical
solution of this problem is of particular interest. The graphical
as well as the algebraic method of solution of such a problem is
illustrated by Problem 62.
ILLUSTRATIVE PROBLEMS
60. In the steelyard shown in Fig. 79 the distance AB = 5 in. Find the
force R acting at B and the distance BC if the steelyard is balanced.
Solution. —
2^ = 72-40-10=0,
•. i2 = 50 1b.
XMb = ^0X5-10XBC = 0,
BC=^ = 20in. I '^«>^-
10
-D
Fig. 79.
61. A load of 1200 lb. is applied to a beam AB as shown in Fig, 80(a).
The left end of the beam is carried by a second beam CD. Find the reactions
on the second beam at C and D.
80
EQUILIBRIUM OF FORCE SYSTEMS
Solution. — The free-body diagrams for the two beams are shown in
Fig. 80(6) and 80(c). The equation XMa=0 becomes
1200X9-/2^X12=0,
/. 7^5 = 900 lb.
There is, then, a load of 900 lb. acting on the beam CD at B. The two reac-
tions Re and Rd may be found by applying either set of equilibrium equations.
Thus, using the equations SMc = 0 and 7^Md = 0, the two reactions may be
found as follows,
i2r,XlO-900X6 = 0,
-i2cXl0+900X4=0,
n
il200 lb.
^ !
^"^.fil-J^J^/
(a)
B
1200 lb.
900 1b.
C Y D
Fig. 80.
////)/J
ib)
i2z) = 540 lb.
i?c = 360 lb.
2000 lb. 2000 lb.
1000 1b.
*-2 >|< 3-
Ri
Fig. 81.
62. A beam 12 ft. long carries three loads as shown in Fig. 81. Find the
reactions at the ends of the beam.
Algebraic Solution. —
SMjs =- 12i2i +1000 X 10-1-2000 X7+2000 X4 = 0,
12i2i = 32000,
/. El =26661 lb.
SMa = 12/^2-1000X2-2000X5-2000X8=0,
12/22 = 28000,
/. 722 = 2333 J lb.
In order to check the results the equation S/^=0 may be applied. Thus,
26661 +2333i = 1000+2000+2000.
EQUATIONS OF EQUILIBRIUM
81
Graphical Solution. — In order to determine the reactions by the graphical-
method, a force and a funicular polygon are constructed as shown in Fig. 82.
Since the forces are in equilibrium the two polygons must close. The three
known forces are represented by AB, BC, and CD. The right reaction will be
represented by DE, the position of the point E being as yet unknown. Obvi-
b
c c
d d
A
B
E
C
e
a a
a ^
b
f ~~^~~^
e __
o _,
^^d
(6)
Scale. 1 in. = 10 ft.
D
^.-" (6)
Scale. 1 in. =4000 lb.
Fig. 82.
ously the left reaction will be represented by EA since the force polygon must
close. The strings oa^ ob, oc, and od of the funicular polygon are drawn par-
allel to the corresponding rays. Since the string oe must intersect the string
oa on ea and since oe must also intersect the string od on de, the position of the
string oe is determined. The direction of the ray OE is now determined
also, since it must be parallel to the string oe. Hence E is the point where a
line drawn through 0 parallel to oe intersects the line AD. The magnitudes
of DE and EA are found, from the diagram, to be 2330 lb. and 2670 lb., which
agree closely with the values found by the algebraic method of solution.
PROBLEMS
63. A beam 14 ft. long carries a concentrated load of 1000 lb. and a uni-
formly distributed load of 200 lb. per linear foot as shown in Fig. 83. If
the weight of the beam is neglected find the reactions at the ends of the beam.
F
1000 lb.
200 lb. per ft.
m
-H^
Rj
Fig. 83.
500 1b.
200.1b.
''zi^
y/^y'^
10-
FiG. 84.
64. The beam shown in Fig. 84 weighs 10 lb. per linear foot. Find the
reactions due to the weight of the beam and the loads shown.
Ans. i2i = 202 1b.; /?2 = 638 lb.
82
EQUILIBRIUM OF FORCE SYSTEMS
66. A bar 8 ft. long is held in equilibrium by the three forces shown in
Fig. 85 and two forces acting along the lines ef and de. Find the magnitudes
and senses of the two forces.
Ans. DE=-\-mO lb. EF= -740 lb.
ymf/////////////^^^^^^
e
J
200 lb. "^
400 lb.
. 420 lb.
Fig. 85.
66. A load of 1000 lb. on the system of pulleys shown in Fig. 86 is held
b}'- a force P. Find the magnitude of P and of the stresses in all cords.
Neglect the weights of pulleys and axle friction.
§ 5. Non-concurrent, Non-parallel Forces in a Plane
50. Equations of Equilibrium. — A system of coplanar, non-
concurrent, non-parallel forces is in equilibrium if the forces of the
system satisfy the equations,
SF, = 0
SF, = 0 ■ , {A)
where x and y denote the coordinate axes and A is any point in the
plane of the forces.
Proof. — If a system of non-concurrent, non-parallel forces in a
plane is not in equiUbrium, the resultant of the system is either a
force having components equal to ^Fx and SFj, or a couple hav-
ing a moment equal to SAf^ (Art. 30). If the equation 2Fi=0 is
EQUATIONS OF EQUILIBRIUM 83
saiisficd the resultant, if a force, must be parallel to the y-a^sT
If the equation XFy = 0 is satisfied, the resultant, if a force, must be
parallel to the x-axis. A force cannot be parallel to both the
X- and y -Sixes and hence, if the first two equations are satisfied,
the resultant of the system cannot be a force. If the equation
SMa = 0 is satisfied, the resultant cannot be a couple. Therefore,
if the forces of the system satisfy the three equations, the force
system is in equilibrium.
Another set of independent equations of equilibrium for a
non-concurrent, non-parallel system of forces in a plane may be
written as follows:
(B)
where x denotes any line or axis in the plane of the forces and A
and B are any two points in the plane, provided that the line AB is
not perpendicular to the a;-axis.
A third set of equihbrium equations for the force system here
considered may be written as follows :
(C)
where A, B, and C are any three non-coUinear points in the plane
of the forces.
It will be left to the student to prove that either set of equa-
tions {B) or (C) are sufficient and necessary to ensure the equihb-
rium of a coplanar, non-concurrent, non-parallel system of forces.
Any one of the above sets of equations, therefore, may be used
to determine the unknown elements of a coplanar, non-concurrent,
non-parallel force system which is in equilibrium, provided there
are not more than three such unknown elements or quantities.
84
EQUILIBRIUM OF FORCE SYSTEMS
ILLUSTRATIVE PROBLEMS
67. A ladder rests with its upper end against a smooth vertical wall and
its lower end on a smooth horizontal floor as shown in Fig. 87. SHpping
is prevented by means of a stop at the lower end. If the ladder weighs
50 lb. and is of uniform cross-section, find the reactions of the wall, the floor,
and the stop, on the ladder.
Solution. — ^A free-body diagram of the ladder is shown in Fig. 87. Since
the wall and floor are smooth, the reactions at these surfaces are perpendicular
to the surfaces and since the ladder has a uniform cross-section, its weight
acts at the center of the ladder. Let the length of the ladder be denoted by I.
The solution is effected by applying the three equations of equilibrium as
follows :
^Ma=RiXI sin 60°-50XZ/2
cos 60° = 0,
.-. 7^1 = 25 cot 6P° = 14.42 lb.
SFx = i?3-i2i = 0,
/. R, = R^ = UA2\h.
sn =7^2-50 = 0.
Fig. 87.
.*. 7^2 = 50 lb.
68. The wall bracket shown in Fig. 88(a) consists of a horizontal member
AB, which is attached to the wall at A by means of a smooth pin, and a rod
C5, which is attached to the member AB at B and to the wall at C by means
4001b
' 1000 U>. ,
A
-4;- — >*<-
400 lb.
(6)
' 1000 lb.
Fig. 88.
of smooth pins. Find the tension, 7", in the rod and the pin reaction, R, at
A if the weights of the members are neglected.
Solution. — A free-body diagram of the member AB is shown in Fig. 88(6)
GRAPHICAL SOLUTION OF A TYPICAL PROBLEM
85
There are three unknown quantities in the force system, namely, T, R, and d.
The three independent equations of equilibrium are,
SMA = 7^Xl2sin30°-1000XlO-400X4 = 0, (1)
.-. T = 1930 lb.,
i:Fx = Rcosd-T cos 30° = 0,
/. 22 cos 0 = 1930 cos 30° = 1670, (2)
XFy=R sin e+T sin 30° -400- 1000 = 0,
.*. 72 sin 0 = 1400 -1930 sin 30° =435 (3)
By solving equations (2) and (3) the following results are obtained,
72 = 1730 lb.
0 = 14° 35'.
PROBLEMS
69. A bar (Fig. 89) leans against a smooth vertical post and rests with its
lower end on a smooth horizontal plane, shpping of the lower end being pre-
vented by the cord as shown. If the weight of the bar is neglected, find the
reactions of the plane at A and of the post at B and also find the tension in the
cord.
Ans. 72^ = 10.9 lb. 72^ = 51.1 lb. T = 32.8 1b.
Fig. 89.
Fig. 90.
70. In Fig. 90 is shown a bell-crank mounted on a smooth pin at O and
subjected to a force of 80 lb. at A as shown. Find the pin pressure at O and
the reaction at B. Solve algebraically and graphically.
51. Graphical Solution of a Tjrpical Problem. — Any of the
problems which involve a balanced non-concurrent, non-parallel
force system in a plane in which there are not more than three
unknown elements may be solved graphically as weU as alge-
86
EQUILIBRIUM OF FORCE SYSTEMS
braically. A graphical method of solution for one typical problem
will here be discussed. In the force system considered all of the
forces will be assumed to be known completely except two, the
action line of one of these two being known and also one point on
the action line of the other. The three unknown elements, then,
are the magnitude of one of the two forces and the magnitude and
the direction of the other.
As an example of such a force system, consider the forces
acting on the beam shown in Fig. 91(a). The unknown ele-
ments are the magnitude of the vertical reaction at the left end
of the beam and the magnitude and the direction of the reaction
at the right end of the beam.
b 6V c^^f
C«)
Fig. 91.
The force and funicular polygons are shown in Fig. 91. In
constructing the force polygon, AB, J5C, and CD are first drawn
to represent the three known forces (Fig. 916). Since the mag-
nitude of the force BE is unknown the location of E is not known,
but it must lie in a vertical hne through D. The rays are then
drawn from 0 after which the funicular polygon is constructed
(Fig. 91a). Since the point iV is the only known point on the
action line of the force EA^ the funicular polygon must be started
at this point. The strings oa, ob, oc, and od are drawn as shown.
Since the string oe must intersect od on de and must also intersect
oa on ea its position is determined. The ray OE must be parallel
to the string oe. Hence E is the point of intersection of a vertical
line through D and a line through 0 parallel to oe. DE then rep-
resents the left reaction, and the right reaction is represented in
magnitude and in direction by EA.
CHOICE OF MOMENT-CENTERS
87
52. Choice of Moment-centers and of Directions of Resolu^
tion. — In applying the three equations of equilibrium to a balanced
force system, of the type here considered, in wliich there are not
more than three unknown elements, the solution of the problem
may frequently be simplified by selecting the moment-centers and
the directions of resolution in a particular way. For example, in
Fig. 92 is represented a portion of a roof truss which is held in
equiUbrium by the five forces shown, of which P and R are known
completely and Fi, Fo, and F3 are unknown in magnitude. By
selecting C, the intersection of the two unknown forces F2 and F3,
as a moment-center and applying the equiUbrium equation
2Mc = 0, the force Fi may be found from the one equation. Like-
wise by choosing D as a moment-center and applying a second
equation of equihbrium, SMzj = 0, the force F3 may be found
directly. Similarly, F2 may be found by selecting A as the
moment-center and applying the third equihbrium equation,
SMa = 0. Thus by the proper selection of moment-centers each
of the three equations involves one unknown quantity only.
V A
F>S
\
-^^^>-2-
Fig. 92.
Fig. 93.
As another example, consider a body which is held in equi-
librium by the six forces shown in Fig. 93, all of which are com-
pletely known except Fi, F2, and Fs, which are unknown in mag-
nitude. The forces Fi and F3 are parallel. By selecting B as the
moment-center, the force Fi may be found from one equation,
namely, 1:Mb = 0. Likewise, 7^3 may be found from the single
equation Silf a =0, where A is the intersection of the two forces
Fi and F2. Further, F2 may be found from the single equation
XFj, = 0 provided that the ^/-direction is chosen perpendicular to
the forces Fi and F3.
88 EQUILIBRIUM OF FORCE SYSTEMS
§ 6. Equilibrium of Trusses, Cranes, and Cables
53. Stresses in Trusses. — Important examples of balanced
coplanar force systems are met in the analysis of the forces in
trusses, cranes, and flexible cables. In determining the stresses
in pin-connected trusses, only those trusses will here be con-
sidered for which the following assumptions may be made :
(1) The members of the truss lie in one plane and hence the
forces acting on the truss and the stresses in the members form a
coplanar force system.
(2) The members of the truss are connected at their ends by
means of smooth pins.
(3) The loads on the truss act only at the pins, that is, at the
ends of the members.
(4) The weights of the members are neglected, since the stresses
due to the weights are small in comparison with the stresses due to
other loads.
For practical purposes, in computing the stresses in many
structures such as roof and bridge trusses, the structures may be
considered to be in agreement with the above assumptions.
According to assumptions (2) and (3), the only forces acting
on any member are the reactions of the smooth pins at the ends of
the member, and since the member is in equilibrium under the
action of these two forces, the two forces must be collinear, and
hence the action Hue of each force must coincide with the axis of
the member. Therefore the stress in each member is a direct or
axial pull or push (tension or compression). Thus in the truss
shown in Fig. 94(a), the member AB is acted on by the pin pres-
sures 7^1, Fi at the ends of the member which cause a compressive
stress in the member (Fig. 946), while the pin pressures F2, F2
which act at the ends of member AC cause a tensile stress in the
member (Fig. 94c). The stress S, then, in any member is numeric-
ally equal to the pressures of the pins at the ends of the member,
as is indicated in the freebody diagrams of the two parts oi AC
(Fig. 94d). The kind of stress (tension or compression) in a mem-
ber, therefore, is represented by arrows which show the directions
of the pressures of the member on the pins. Thus, a compressive
stress is represented by arrows along the member directed towards
the ends of the member, etc. Thus, the stress in member AC (Fig.
METHOD OF JOINTS
94a) is shown to be tension and the stress in member AB is shown
to be compression. Another convention commonly used to indi-
cate the kind of stress in a member is to represent a tensile stress
by a plus sign and a
compressive stress by a ^^ P- ^^ p*
negative sign. In the
analysis of the stresses
in a pin-ended structure
two rather distinct meth-
ods may be used, name-
ly, the method of joints
and the method of sec-
tions.
54. Method of Joints.
— By the method of
joints the stresses in
the members of a truss
are found by the solution
of the concurrent force
systems which act at
the pins, and hence not
more than two unknown
pin pressures at any joint or pin can be determined. For
example, in the truss shown in Fig. 94(a) the pin at A is in
equilibrium under the influence of three forces (Fig. 95a), namely,
the reaction Ri, which may be found if the dimensions of the truss
and the loads carried by the truss are known, and the two unknown
forces (stresses) Fi and F2 which
the members AB and AC exert
on the pin. By using the two
equations of equilibrium for a
R,-^
A
I
C
(6)
■^F.
system of concurrent forces in a
plane the two unknown forces may
Fig. 95. be found. The pin at C may now
be considered. This pin is in
equilibrium under the influence of three forces (Fig. 956) namely,
the stress F2 in the member AC which has just been determined,
and the two unknown stresses F3 and F4, in the members CB and
CE. These two unknown stresses may be determined from the
two equations of equilibrium. The pin at B may next be con-
90 EQUILIBRIUM OF FORCE SYSTEMS
sidered, and so on until the stresses in all of the members have
been determined. As already noted, in using this method the
joints must be taken in such an order that there are not more than
two unknown stresses at the joint considered, since there are only-
two equations of equiUbrium for a concurrent force system and
hence only two unknown stresses may be determined. In con-
sidering the equilibrium of the forces at any joint the senses of the
two unknown forces are not always evident. When such is the
case the forces may be assumed to be directed away from the joint,
that is, the stresses may be assumed to be tensions. If, after
applying the equations of equilibrium and solving for the two
unknown forces, the sign of a force is found to be positive it is
evident that the assumed sense of the force is correct, whereas a
negative sign indicates that the sense of the force is opposite to
that assumed. In most problems, however, the sense of each
of the unknown forces is evident from inspection. The deter-
mination of the stresses in a truss by the method of joints often
requires a large number of calculations with a corresponding chance
for error. Thus, if the stress in a single member near the center
of the truss is required it is necessary to start at the end of the
truss and consider the equilibrium of the joints in order until a
joint is reached which involves that particular member. By the
method of sections (Art. 55) the stress in a single member may
frequently be found by use of a single equation.
55. Method of Sections.— In determining the stresses in the
members of a pin-connected truss by the method of sections a
section is passed through the truss so that it cuts not more than
three members in which the stresses are unknown. The part of
the truss on either side of this section is then treated as a free
body in equiUbrium under the action of the known external forces
which act on that part and the forces (stresses) which the members
of the other part exert on the part considered. It will be noted
that the method of sections involves a non-concurrent force system
and hence there are three equations of equiUbrium from which
three unknown forces may be found. Consequently the section
must not cut more than three members in which the stresses are
unknown. As a rule any one of these three unknown forces
(stresses) may be found by usmg one equilibrium equation as dis-
cussed in Art. 52. For example, consider the truss shown in
Fig. 94. In order to determine the stress in the member DG con-
METHOD OF SECTIONS
91
sider the section which cuts the members DF, DG, and EG. The
free-body diagram of the portion of the truss to the left of this
section is shown in Fig. 96(a). This part of the truss is in equi-
hbrium under the action of six forces; namely, Ri, Pi, and P2,
which are assumed to be known, and the unknown stresses in the
members DF, DG, and EG. The stress (Fq) in the member DG
may be obtained from the equation SFy = 0, since neither of the
other two unknowns will appear in this equation. Similarly the
stress (F5) in the member DF may be obtained from the single
equation 'EMg=0 and the stress {F7) in the member EG may be
obtained from the single equation XMd = 0.
1 Fig. 96.
It is important to note that the forces F5, Fq, and F7 are
regarded as internal forces (stresses) when the equilibrium of the
truss as a whole is considered, but that they are regarded as
external forces when the part of the truss on one side of the section
is considered. In considering the equilibrium of the part of the
truss on the left of the section (Fig. 96a) only the external forces
which act on the left part are involved and hence the stresses in
the members of the left part that are not cut by the section do
not affect the values of F5, Fq, and F7. Thus the left part may
be thought of as a solid block as indicated in Fig. 96(6) on which
the forces F5, Fq, and F7 act together with Ri, Pi, and P2, the six
forces holding the block in equilibrium.
ILLUSTRATIVE PROBLEM
71. Determine the stresses in the members of a Howe truss when loaded
as shown in Fig. 97(a).
Solution. — The stress in any member will be denoted by the same letters
as the member itself. Thus, the stress in member AB will be denoted by
AB. By considering the equilibrium of the truss as a whole, the reactions
d2
EQUILIBRIUM OF FORCE SYSTEMS
40,000 lb.
20,000 lb.
20,000 lb.
CE
ie)
EG
Fig. 97.
METHOD OF SECTIONS 93
Ri and R2 are found to be 40,000 lb. To find the stresses in members AB and
AC the joint method is used. A free-body diagram of joint A is shown in
Fig. 97(6). By applying the equations of equilibrium for a concurrent force
system, the stresses in the two members are found as follows:
SMb= -40,000X20H-ACX30 = 0,
.-. AC = iO'Q^X^g = 26,687 lb. (tension)
oO
XFy=-AB COS «/) +40,000 = 0.
.-. A5 = ^^— = 40,000 sec 0 = 48,000 lb. (compression)
cos 0
To find the stresses in members BC and BD let a section aa be passed
(Fig. 97c). The forces acting on the part of the truss to the left of the section
form a non-concurrent force system. But since the stress in member AC is
already known only two of the three equations of equilibrium are needed.
Thus,
2M c = - 40,000 X 20 -\-BD X 30 = 0,
/. BD = ^^'^^^^^^ = 26,667 lb. (compression)
XFy = 40,000 - 20,000 -BC = 0,
:. BC = 20,000 lb. (tension)
In a similar manner the stresses in CD and CE may be found by the
method of sections. By considering the part of the truss to the left of sec-
tion 66 (Fig. 97d), the stresses in the two members are found as follows:
Sikfi) = -40,000 X40 +20,000 X 20 +C^ X 30 = 0,
.-. CE = 1^?^^ = 40,000 lb. (tension)
SF,/ = 40,000- 20,000 +CD cos 0=0,
/. CD= -20,000 sec </>= -24,000 lb. (compression)
It will be noted that all of the stresses except CD (Fig. 97rf) are assumed
to act in the correct directions, and hence are found to be positive whether
they are tensile or compressive. The stress in CD is assumed to be tension
and hence the negative sign indicates that it is compression.
In considering the equilibrium of the forces which act on the pin E (Fig.
97e), it is evident that the stress in DE is zero, and that EC equals EG. Fur-
ther, since the truss is symmetrical with respect to the center line DE and the
loads are also symmetrical with respect to this line, it is obvious that the
stresses in the members of the right half of the truss are equal to the stresses
in the corresponding members of the left half.
94
EQUILIBRIUM OF FORCE SYSTEMS
PROBLEMS
72. Find the stresses in the members of the Warren truss shown in Fig. 91
Am. AB=- ^Q20\h. D^=-34601b.
4000 lb.
2000 1b
Fig. 98.
73. Find the stresses in the members of the Pratt truss shown in Fig. 99
in which the loads P are each 10,000 lb.
Ans. AB = -30,000 lb. BC^ +10,000 lb. BD = -26,670 lb.
Fig. 99.
74. Find the stresses in the members of the Howe truss shown in Fig. 100.
Ans. A5= -12.700 lb. BC= -3000 lb. ^D= -9000 lb.
Fig. 100.
GRAPHICAL ANALYSIS OF TRUSSES 95
75. In the Fink truss shown in Fig. 101 BC is perpendicular to AD and K
is the mid-point of AD. Find the stresses in the members.
Ans. BC=- 1730 lb. AB = -6000 lb.
2000 lb.
2000 lb. 2000 lb.
Fig. 101.
56. Graphical Analysis of Trusses. — The graphical method of
analysis of framed structures is sometimes simpler than the
algebraic method. This statement is true particularly when the
form of the structure is such that a considerable amount of cal-
culation is necessary to determine the directions and moment-arms
of the forces involved. The graphical method consists essentially
in constructing the force polygons for the concurrent forces at the
joints and superimposing these polygons. The method will be
explained in detail with reference to the Pratt truss shown in
Fig. 102(a). It will be assumed that the loads carried at the upper
panel points are equal. It is convenient to use the Bow system of
notation (Art. 7). In this system the regions on either side of
the action line of a force (either external or internal) are denoted
by numbers or letters as shown in Fig. 102(a) and the force is
denoted by the two mmibers or letters adjacent to the action line
of the force. For example, the left reaction is denoted by Xi-Y;
the load at the upper panel point at the left end of the truss by
Xi--X'2; the stress in the diagonal in the second panel by 2-3;
the stress in the lower cord at the left end of the truss by
Y-1; etc. The order of the numbers or letters denoting an
external force, or an internal stress in any member, may be reversed.
Thus the stress in the diagonal in the second panel may also be
denoted by 3-2. The force polygon shown in Fig. 102(6) is con-
structed as follows: The load line X1X2X3X4X5X6 is first laid off.
Since the loads are symmetrical, the reactions will be equal,
and hence Y will be midway between the points Xi and Xq.
96
EQUILIBRIUM OF FORCE SYSTEMS
Consider now the three concurrent forces at the lower left-hand
panel point. Of these three forces the left reaction, Y-Xi, is com-
pletely known and the directions of the other two, Xi-1 and
1-F, are known. Since these three forces are in equilibrium,
their force polygon must close. Hence, to determine the mag-
nitudes of the last two forces a Hne is drawn in Fig. 102(6)
through Xi parallel to Xi-1, and a line is drawn through Y par-
allel to 1-Y. These lines intersect in the point 1, and the mag-
nitudes of the stresses are indicated by the lengths of the lines
Xi-1 and 1-Y in the force polygon. It will be noted that the
pressure of the member Xi-1 on the pin at the lower end is down-
ward to the left and hence the member is in compression.
1
' Xg )
- X3
. X4 .
r X5 ,
ii
i
^
^
^
H
(a)
Fig. 102.
As soon as the stresses in the various members are found
arrow-heads should be placed on the members indicating whether
the stresses are tensions or compressions. The second lower panel
point should next be taken. The stress in the member F-1 is
now known and in order that the polygon for this joint shall close
it is evident that the point 2 must coincide with 1. Therefore
the stress in 1-2 is zero and the stress in 2-F is the same as the
stress in Y-1. The first upper panel point can now be taken and
the stresses in X2-S and 3-2 may be found. The forces at the
remaining joints may be found in a similar manner. The student
should follow through the construction (Fig. 1026) and build up
the complete polygon.
GRAPHICAL ANALYSIS OF TRUSSES
97
ILLUSTRATIVE PROBLEM
76. Find the stresses in the members of the truss shown in Fig. 103(a).
The members Xi-1 and 2-3 are parallel and the members Xi-2 and F-l
are parallel.
+13500
Fig. 103.
Solution. — The stresses are determined from the force polygon shown in
Fig. 103(6), The magnitudes of the stresses are shown on the members
in Fig. 103(a), the plus or minus sign indicating whether the stress in the
member is tension or compression.
PROBLEMS
77. The Fink truss shown in Fig. 104 is subjected to wind loads as indi-
cated, the loads being perpendicular to the upper chord AD. The truss
Fig. 104.
rests on a smooth plate at the left end and hence the reaction at that end is
vertical. Find the magnitude and direction of the reaction at the right end
of the truss and the stresses in the members. Use the graphical method.
98 EQUILIBRIUM OF FORCE SYSTEMS
78. Find, by the graphical method, the stresses in the members of the
truss shown in Fig. 105, the value of P being 10,000 lb.
'P
P
20 = 120^
'P
ill
Fig. 105.
57. Cranes. — It was stated in Art. 53 that in the analysis of the
forces which act on the members of a crane balanced coplanar
force systems are involved. The loads on cranes, unlike those on
trusses, are not applied at the pins only. That is, assumptions
(2), (3) and (4) of Art. 53 may not always be made in the anal-
ysis of the forces acting on cranes. The force system acting on
a member of a crane, therefore, includes forces other than the
two pin pressures at the ends of the member, and hence the stress
in a crane member, in general, is not a direct axial tension or com-
pression as is the case in a truss. In order to calculate the stress
in a crane member a knowledge of the subject of Strength of
Materials is needed, and consequently the stresses in crane mem-
bers will not, in general, here be foimd. The main problem here
considered in connection with cranes, therefore, is that of deter-
mining only the forces acting on the various members of the
crane. Since the stresses in the members of a crane are not single
forces directed along the axes of the members, the method of
sections as used in the preceding articles cannot be employed.
The general method of procedure in the analysis of the forces
acting on crane members will be first to consider the equihbrium
of the structure as a whole and determine the external reactions,
and then to consider the equilibrium of the various members
separately and determine the forces (pin pressures, etc.) which
act on the separate members.
CRANES
99
ILLUSTRATIVE PROBLEM
79. In the crane shown in Fig. 106 the post AD weighs 1200 lb. and the
boom BE weighs 1000 lb. The member CE is a cable the weight of which
will be neglected. The reaction at D will be assumed to be horizontal. Find
the reactions at A and D, the pin reaction at B, and the tension in the cable CE.
^////Mm
_D
D
'\
12'
^y\
' C
y
yy
/^
"c
i
1200 lb.
>/i
^
1
6000
lb.
y
1000 lb.
"i
■
(a)
Ax^
A
•
KA
7
/
ky
Fig. 106.
Solution. — First consider the equilibrium of the entire crane, a free-body
diagram of which is shown in Fig. 106(a). The reaction at D will be denoted
by D. The reaction at the pin A will be denoted by A and, for convenience,
will be resolved into its horizontal and vertical components which will be
denoted by ^a; and Ay. The unknown reactions are obtained by applying
the equations of equilibrium as follows:
SMa = 20Z) - 6 X 1000 - 12 X 6000 = 0,
... 0 = 1^ = 3900 lb.
XFx = Ax-D = 0,
.-. D = Ax = 3900 lb.
2Fy = Ay - 1200 - 1000 - 6000 = 0,
/. Ay = 8200 lb.
100
EQUILIBRIUM OF FORCE SYSTEMS
Whence,
and,
A = VAx^+Av^ = V(3900)2+(8200)2 = 9075 lb.
8200
6 A = tan
3900
64° 35'.
Next consider the equilibrium of the member BE. The forces acting on this
member are four in number; namely, the tension in CE which will be denoted
by CE, the load of 6000 lb., the weight of the member, and the pressure of the
pin at B on the member, as shown in the free-body diagram (Fig. 1066). The
last force is resolved into its horizontal and vertical components as shown in
the free-body diagram. By applying the equations of equilibrium the un-
known quantities may be found as follows:
2Mb = 9XC^-6X1000-12X6000 = 0,
,. CE=I^ = 8667 1b.
i:Fx=Bx-CE=0,
.*. Bx = CE = SQQ7lh.
XFv = By- 1000 - 6000 = O.
;. By = 7000 lb.
^ = V5a;2+Bv2 = V(8667)2+(7000)2 = ll,1001b.
^B=tan-i|500^3go55,
8667
Whence,
and,
80. Determine the forces acting on the members of the crane shown in
Fig. 107(a). Neglect the weights of the members.
1000 lb.
Fig. 107.
CRANES 101
Solution. — Consider first the equilibrium of the entire crane, a free-bodj^
diagram of which is shown in Fig. 107(a). By applying the three equations
of equilibrium, the unknown quantities may be found. Thus,
2Ma =20 XC^-30X 1000 = 0.
.-. C^ = 5^^ = 1500 lb.
XFx = Ax-CE sin S0° = 0,
:. Ax = C^sin30° = 7501b.
i:Fy=Av-lOOO-CE cos 30°=0,
.*. Ay = 1000 +1300 = 2300 lb.
Whence,
and.
A = V(750)2 + (2300)2 = 2420 lb.
0A=tan-i^ = 71°55'.
Consider next the equilibrium of the member BH, The forces acting on
BH are shown in Fig. 107(6) . From the equations of equilibrium the unknown
quantities may be found as follows :
2M^ = 10.6XCD-30X1000=0,
/. CD = ??^ =2830 lb.
lO.o
^Fx = Hx-CD cos 45° = 0,
/. i7x = 2830 X. 707 = 2000 lb.
SFy= -Hy+CD cos 45° -1000 = 0,
*. H.,= -1000+2830X.707 = 10001b.
H = V(2000)2+(1000)2 = 2240 lb.
PROBLEMS
81. In the crane shown in Fig. 108 the post AE weighs 1600 lb. and the
member CH weighs 1200 lb. The remaining members are cables the weights
of which may be neglected. The member CH passes through a slot in the
post AE. Determine the external reactions, and the tensile stresses in the
members BC, CD, and DH. Assume the reaction at E to be horizontal and
that there is no reaction between CH and AE
Whence,
and,
102
EQUILIBRIUM OF FORCE SYSTEMS
82. The crane in Fig. 109 is in equilibrium under the action of the two
loads shown and the reactions at B and A, the reaction at B being horizontal.
Find the reactions at A and B and the tension in the member DF. Find also
the pressure of the pin at C on the member CF.
Ans. 5 = 5200 lb. D/^ = 15400 lb.
I
'^yy////A//J//
6000 lb.
Fig. 108.
Fig. 109.
58. Flexible Cables. — In the following two articles the equi-
librium of flexible cables or cords will be discussed. A cable is
said to be perfectly flexible when it can offer no resistance to bend-
ing. A flexible cable, then, can transmit a stress only along its
axis, that is, the stress at any point of a flexible cable is tangent
to the curve assimied by the cable. Although physical cables
and cords are not perfectly flexible the resistance they offer to
bending is generally so smaU that it can be neglected without
serious error. In the discussion of cables it wiU be assumed that
the cables are perfectly flexible and inextensible.
59. The Parabolic Cable. — If a flexible cable is suspended
from two points and carries a load that is distributed uniformly
horizontally (Fig. 110a), the curve assumed by the cable is a para-
bola, as will presently be shown. In the present discussion the
points from which the cable is suspended will be assumed to be in
the same horizontal plane. An example of a cable carrying a load
which closely approximates that above indicated is the cable of a
suspension bridge, since the weight of the roadway is uniformly
distributed horizontally and the weights of the cable and hangers
are small in comparison with the weight of the roadway and
THE PARABOLIC CABLE
103
therefore may be neglected. Another example is that of— ar
tightly stretched cable (that is, one in which the sag is small as
compared with the span) which carries no load except its own
weight, as for example, the cable of an electric transmission line,
a telegraph wire, etc. In this case the load carried by the cable
(its weight) is distributed uniformly along the curve assumed by
the cable, but since the sag is small the horizontal projection of an
arc of the curve is approximately equal to the length of the arc,
and hence the load is distributed approximately uniformly in the
horizontal direction.
Fig. 110.
In the solution of problems involving the parabolic cable, use
is made of the equation of the curve assumed by the cable (para-
bola) and of equations which express relations between the span,
sag, length of the cable, tension, etc. In order to determine the
equation of the parabola a portion AB oi the cable wiU be con-
sidered as a free body (Fig. 1 106) . A , the lowest point of the cable,
will be taken as the origin of coordinates and the tension at this
point will be denoted by H. The tension at any point, B, will be
denoted by T. The portion of cable AB, then, is in equilibrium
under the action of the three forces Hj T, and the vertical load
wx which acts through the point D midway between A and C.
Since these three forces are in equilibrium they must be concurrent,
104 EQUILIBRIUM OF FORCE SYSTEMS
and hence the action Una of T passes through D, The equa-
tions of equihbrium are:
1:Fx=T co&a-H = Q, (1)
i:Fy=T ^ina-wx = Q (2)
By dividing (2) by (1), the equation obtained is,
wx
But,
Hence,
tan a— jj ,
tan a = — .
X
2y _wx
That is,
.^g. .......... (3)
The curve, then, is a parabola with its vertex at A and its axis
vertical. By squaring and adding (1) and (2) and extracting the
square root of each side of the resulting equation, the following
expression for T is found,
T=VlP+w^ (4)
In applying the above equations we are concerned with the tension
at the point of support since at this point the tension is a maximum.
Hence, if the span be denoted by a and the maximum value of y
(that is, the sag) by /, equations (3) and (4) reduce to
/=S • (^)
T = iwayjl+^„ (6)
in which T represents the tension at the points of support.
The length of the cable will now be determined in terms of the
span and sag. In any curve the length of an arc is obtained from
the equation
" ^)-
T^ .• /o\ dy ^^
From equation (3), -r^-jr-
and
/^l
THE PARABOLIC CABLE 105
Hence, if the length of the cable be denoted by I, we have
■■f^'
z=2i ^n+^dx.
Substituting for H from equation (5), this equation becomes
-^^/^
1 = 2 \ ^ll + ^^dx.
The exact expression for Z, obtained from this integral, involves a
logarithmic function and is difficult to apply. A simpler expres-
sion for I may be obtained by expanding the expression under the
integral into a series and integrating the series term by term.
This method leads to the following equation,
'-Mff-m- ■■] ">
Since the sag ratio f/a is generally small, the series converges
rapidly and it is sufficient in most practical computations to use
only the first two or three terms of the series to obtain a close
approximation to the value of I.
ILLUSTRATIVE PROBLEM
83. The horizontal load carried by each cable of a suspension bridge is
1000 lb. per ft. The span of the bridge is 800 ft. and the sag is 50 ft. Deter-
mine the tensions at the ends and at the middle of the cable and also find the
length of the cable.
Solution. — From equations (5) and (6),
and,
g= oC.n =1,600,000 lb.
T = ^Xl000X800>/l + j^^^^^ = l,650,000lb.
The length of the cable may be determined by using equation (7) . Thus,
106
EQUILIBRIUM OF FORCE SYSTEMS
PROBLEMS
84. A telegraph wire weighing 0.1 lb. per foot is stretched between
two poles 150 ft. apart. The tension in the wire at the insulators (which
are in the same horizontal plane) is 500 lb. Find the sag, assuming
that the weight of the wire is uniformly distributed horizontally. Find also
the length of the wire.
Ans. /= 0.562 ft. I = 150.005 ft.
85. Each cable of a suspension bridge carries a load of 1200 lb. per foot
uniformly distributed along the horizontal. The span is 1000 ft. and the sag
is 50 ft. Find the maximum stress in the cable and the length of the cable.
86. A cable 100 ft, in length is suspended from two points in a horizontal
plane which are 99 ft, apart. If the cable carries a load that is uniformly
distributed along the horizontal what is the sag of the cable?
^ns./= 6.10 ft.
60. The Catenary. — The curve assumed by a flexible cable of
uniform cross-section which is suspended from two points, and
which carries no load except its own weight (Fig. Ill a), is called a
Fig. 111.
catenary. The load which causes a cable to assume the form of
a catenary, then, differs from that which causes the form of a
parabola in that the load is distributed uniformly along the cable
in the former case, whereas in the latter case the load is distrib-
uted uniformly horizontally.
The discussion of the catenary is of practical importance only
TOE CATENARY i07
for cables in which the sag ratio is large, since for a small ~sag^
ratio the curve assumed by a cable may be regarded with small
error as being a parabola, as discussed in the preceding article.
In order to determine the equation of the catenary and also
to derive certain important relations between such quantities as
the sag, span, length of cable, tension, etc., the equilibrium of a
portion, OA, of the cable (Fig. 1116) will be considered, 0 being the
lowest point of the cable and A any other point. The point 0
will be taken as the origin of coordinates, the weight of the cable
per unit of length will be denoted by w, and the length of the arc
OA will be denoted by s. The portion, OA, of the cable is in
equilibrium under the influence of three forces, namely, the ten-
sion H at the point 0, the tension T at the point A, and the weight
ws. The angle which T makes with the horizontal will be denoted
by 6. The equations of equilibrium for the concurrent force
system are,
^F^=T cos d-H = 0, .'. Tcosd = H (1)
i:Fy=T sin d-ws = 0, /. T sin d = ws (2)
From (1) and (2) we have,
Hence,
ws s ' H
isind = -rr = -' where — = c (a constant).
He w
du
s = c tan d or 8 = 0-^ (3)
This equation is the intrinsic equation of the catenary. The
cartesian equation will now be found. In any curve.
Hence, from (3), the following equaton is obtained,
ds _ / c^_ Vs^+c^
dy \ s^ s '
Therefore,
, _ sds
Integrating,
y+A=V^+^.
108 EQUILIBRIUM OF FORCE SYSTEMS
If now the origin is transferred to 0', where 00' = c, then y = c when
s=0 and hence A=0. The last equation, therefore, becomes,
y=Vs^T^. (4)
Eliminating y from (3) and (4),
, cds
ax = —==
Integrating this equation,
x-\-B = c\ogc (s+Vs2+c2).
Since s = 0 when x = 0, B = c loge c and hence the last equation
becomes,
, s+Vs2+c2 y-{-s _.
X=^c\0ge = cl0ge^^ . ... (5)
c c
Equation (5) can also be written in the form,
x_
Vs2+c2+s = ce^ (6)
By inverting each side of (6) and rationahzing the denominator of
the left side, the following equation is obtained,
_i
Vs^-\-c^-s = ce ' (7)
Adding (6) and (7) and using (4),
y = o(^''—^ ^) = ccosh-. (8)
^ c
This is the cartesian equation of the catenary. Subtracting (7)
from (6) we have,
r - -- r
s=7:(e'^ — e '^)=csinh- (9)
2 c
Squaring and adding (1) and (2) we have,
Hence.
T=wy (10)
THE CATENARY 109
Summarizing, then, the following important properties of the"
catenary may be stated :
(1) The horizontal component of the stress at any point is con-
stant and equal to wc.
(2) The vertical component of the stress at any point is equal
to ws.
(3) The total stress T at any point is equal to wy.
In engineering problems which involve the catenary we are
concerned particularly with the tension at the points of support,
since at these points the tension is a maximimi. Hence, in the
above formula, T will be regarded as the tension at the points of
support and the values of x, y, and s will be regarded as the values
of the variables at these points. Therefore, if the length of the
cable be denoted by I, the span by a, and the sag by /, then the
values of x, y, and s in the above equations become ^, /+c, and h,
respectively.
It may be noted that when the sag of the catenary is small the
curve very closely approximates a parabola, since the load is
approximately uniform horizontally. The formulas of Art. 59
are generally used when the sag is small, since they are much easier
to apply and the results obtained are sufficiently accurate for prac-
tical purposes. When the sag is large as compared with the span,
however, the above formulas should be used.
Since the relations between the quantities as expressed by the
above equations are complicated, many of the problems which
involve the catenary can be solved only by trial.
ILLUSTRATIVE PROBLEM
87. A cable weighing 4 lb. per foot is stretched between two points in the
same horizontal plane. The length of the cable is 600 ft. and the tension at
the points of support is 2000 lb. Find the sag and also the distance between
the points of support.
Solution. — From equation (10),
,=1=2^=500 ft.
" w 4
From equation (4),
c = V^^:ii^=V(500)2- (300)2=400 ft.
Hence, / = i/ - c = 500 - 400 = 100 f t .
110 EQUILIBRIUM OF FORCE SYSTEMS
From equation (5),
V;q:^_l_s \/(3oo)2+(400)2+30o
x=c loge ^ -=400 loge -^ ^ '-^
Hence,
=277.2 ft.
a = 2a; = 554.4 ft.
PROBLEMS
88. A cable 100 ft. long is suspended between two points which are in
the same horizontal plane and 80 ft. apart. What is the sag at the mid-point
of the cable?
Am. /= 26.54 ft.
89. A cable weighing 2 lb. per foot is stretched between two points in the
same horizontal plane which are 150 ft. apart. If the sag is 5 ft. what is the
length of the cable and the tension at the points of support.
Ans. ^ = 150.44 ft. 7^ = 1130 lb.
§ 7. Concurrent Forces in Space
61. Equations of Equilibrium. — A non-coplanar, concurrent
system of forces is in equilibrium if the algebraic sums of the com-
ponents of the forces along any three non-coplanar lines through
the point of conciu-rence of the forces are equal to zero. As a
matter of convenience the three hues will be taken as a set of
rectangular axes through the point of concurrence, in which case
the equations of equilibrium may be written :
Proof. — The resultant of a concurrent system of forces in
space, if not in equihbrium, is a force (Art. 32). In order to sat-
isfy the equation SFa;=0, the resultant, if there be one, must He in
the ?/2:-plane. In order to satisfy the equation XFy = 0, the result-
ant must lie in the X2;-plane, and in order to satisfy the equa-
tion 2^2 = 0, the resultant must he in the xy-plsme. It is
impossible for a force to He in the three planes simultaneously and
hence, if the forces of the system satisfy the above equations, the
resultant cannot be a force and, therefore, the system must be in
equiHbrium.
EQUATIONS OF EQUILIBRIUM
111
ILLUSTRATIVE PROBLEM
90. The wall bracket (Fig. 112a) is composed of two flexible cables, AC
and BC, and a stiff rod, DC, which is pin-connected at D and C. The points
A, B, and C lie in a horizontal plane and A, B, and D lie in a vertical plane, D
being vertically beneath E, the mid-point of AB. Find the stresses in the
three members due to the 100-lb. load shown.
X
oiy
tan 0C--\
sin a= ^- =447
y^ (6)
cos a=^-.894
Fig. 112.
Solution. — The pin C is in equilibrium under the action of the 100-lb. load
and the reactions of the three members, these reactions being equal to the
stresses in the corresponding members. The free-body diagram of the pin C
is shown in Fig. 112(a). By selecting axes as indicated and applying the
equations of equilibrium, the following equations are obtained :
2F.
DC =
DC cos 45° -100:
100
0,
cos 45'
141.4 lb.
XFz = AC sin a— BC sin q: = 0,
AC = BC,
2Fx = i)C cos 45'
DCcos45° = 2ACcos«,
DC cos 45
■AC cos a— BC COSa;=0
.-. AC =
2 COS.
141. 4X .707
2X.894
55.9 lb.
Hence there is a compressive stress of 141.4 lb. in the rod DC and a tensile
stress of 55.9 lb. in each of the cables BC and AC.
112
EQUILIBRIUM OF FORCE SYSTEMS
PROBLEMS
91. A weight of 150 lb. is suspended from hooks at points A, B, C, oi &
ceiling as shown in Fig. 113. AD, BD, and CD are cords each 10 ft. in length.
Find the stresses in the cords.
Ans. AD = 71. 68 lb. BD
= 18.08 lb. CZ) = 67.241b.
92. Fig. 114 represents a stifT-
leg derrick. The member AC lies
in the a:2/-plane and the members
BD and BE lie in vertical planes
making angles of 45° with the
xy-plajie. The weight carried is
such as to produce a tensile stress
of 5000 lb. in the member BC.
Find the stresses in the members
AC, BD, and BE.
Ans. AC = 5670 lb.
BE = BD = ^910 \h.
150 lb.
Fig. 113.
Fig. 114.
§ 8. Parallel Forces in Space
62. Equations of Equilibrium.— A system of parallel forces in
space is in equilibrium if the algebraic sum of the forces is zero,
and if the algebraic sum of the moments of the forces with respect
to each of two non-parallel lines is equal to zero, provided that
neither one of the lines is parallel to the forces of the system. It will
be convenient to select a set of rectangular axes so that one of the
EQUATIONS OF EQUILIBRIUM
113
axes (the 2;-axis, say) is parallel to the forces. If the axes are so
selected the equations of equilibrium may be written :
SAf,=0,
Proof. — The resultant of a system of parallel forces in space
which is not in equiUbrium is either a force or a couple (Art. 35).
If 2/^ = 0, the resultant cannot be a force. If the resultant is a
couple it must lie in a plane parallel to the xz-plsme in order to
satisfy the equation '^Mx = 0, and in order to satisfy the equa-
tion SMi, = 0 it must lie in a plane parallel to the yz-plsme.
A plane, however, cannot be parallel to both the xz- and yz-
planes, and hence, if the two moment equations are satisfied the
resultant cannot be a couple. Therefore if the forces of the
system satisfy the three above equations, the force system is in
equilibrium.
ILLUSTRATIVE PROBLEM
93. In Fig. 115, ABC represents a triangular plate, the sides of which are
each 2 ft. in length. It is held in a horizontal position by vertical cords at the
three vertices. A weight of 200 lb is suspended from the point E which lies
on the median AD, the distance DE being 6 in. Find the stresses in the
cords neglecting the weight of the plate.
Solution — The stresses may be found from the equilibrium equations as
follows:
XMx = TsX2 sin 60°-200Xi=0,
100
•. T3 = ,
^ = 57.7 lb.
2 sin 60'
XMz = T2Xl-TiXl=0,
.*. T2 = Tu
27^ = ^1+^2 + 7^3-200=0,
^1+7^2 = 200-^3,
27^1 = 142.3,
/. Ti = T2 = 71.15lb.
Fig. 115.
114 EQUILIBRIUM OF FORCE SYSTEMS
PROBLEMS
94. A square table weighing 50 lb. stands on four legs at the mid-points of
the sides. Find the greatest weight that can be placed on one corner of the
table without causing it to overturn.
95. A uniform circular plate weighing 200 lb. is supported in a horizontal
position at three points on its circumference. Find the reactions at the sup-
ports if the points divide the circumference into arcs of 90°, 135°; and 135°.
§ 9. Non-concurrent, Non-parallel Forces in Space
63. Equations of Equilibrium. — A system of non-coplanar,
non-concurrent, non-parallel forces is in equilibrium if the alge-
braic sum of the components of the forces in each of three direc-
tions is equal to zero and if the algebraic sum of the moments of
the forces with respect to each of three axes is equal to zero, pro-
vided that the directions of resolution are so chosen that lines
drawn through any arbitrary point in tbese three directions are
not coplanar, and that the moment axes do not lie in a plane, and
that no two of them are parallel. It will be convenient to take
the coordinate axes for the axes of resolution and for the moment
axes, in which case the equations of equilibrium may be written
as follows:
Proof. — The resultant of a non-concurrent, non-parallel system
of forces in space is, in general, a force and a couple (Art. 41).
If the first three equations are satisfied the resultant force must
vanish and if the last three equations are satisfied the resultant
couple must vanish. If, therefore, the forces of the system
satisfy the six equations the force system is in equilibrium.
ILLUSTRATIVE PROBLEM
96. Fig. 116(a) represents a windlass used in hfting heavy weights. The
end bearings will be regarded as smooth and the force P apphed to the crank
will be assumed to be perpendicular to the axis of the cylinder and also per-
pendicular to the crank. Find the value of P required to hold the 450-lb
EQUATIONS OF EQUILIBRIUM
115
weight and also find the reactions at the bearings, assuming that the crank is
incUned 30° to the vertical.
450 1b.
Solution. — ^The coordinate axes will be taken as shown in the figure.
There are four forces acting on the windlass, namely, the weight of 450 lb.,
the force P, and the reactions at the bearings. Since the bearing reactions
are unknown in direction as well as in magnitude it will be convenient to resolve
them into horizontal and vertical components, H , Vi and Hzj V2, as indicated
in the figure. Applying the equations of equilibrium to the system of forces
acting on the windlass we have,
i:Fv = P cos SO°+Hi+H 2=0
2/^2 = F1 + F2-P sin 30°-450 = 0. . .
SMx = 15P -450X4 = 0
2Mj, = Psin 30° X 62 +450X30 -50 Fi=0.
2Ms = 50i/i+Pcos30°X62=0. . . .
The solution of these equations gives the following values:
P = 120 lb.
^-1= -128.8 lb.
7i = 344.4 lb.
ff 2 = 24 .92 lb.
72 = 165.6 lb.
(1)
(2)
(3)
(4)
(5)
116
EQUILIBRIUM OF FORCE SYSTEMS
GENERAL PROBLEMS
97. The pressure between the rubbing surfaces of the friction clutch
shown in Fig. 117 is 15 lb. per square inch normal to the surfaces. What
force P on the bell crank is required to produce this pressure?
Ans. P = 196 1b.
^$s?$>^^^;^??ss:^;^^^^;^^^^^^^^^
Fig. 117.
98. A body weighing 20 lb. is held in equilibrium by four cords as shown
in Fig. 118. What are the stresses in cords A, B, and C?
^ns. ^=23. lib. B = 11951b. C = 23 1 lb.
99. A sphere which weighs 40 lb. is held on a smooth inclined plane by
means of a cord which is attached to a ceiling as shown in Fig. 119. Deter-
mine the pressure of the plane against the sphere and the tension in the cord.
yxAy^xyy^yyx^yy^^^^y^
Fig. 120.
100. A body weighing 60 lb. is held in equilibrium on a smooth surface by
two cords which pass over frictionless pulleys and carry suspended weights
of 30 lb. and 50 lb. as shown in Fig. 120. Find the reaction of the surface on
the body and the angle d. Ans. R = 20 lb. 6 = 53° S' .
101. The pillar crane shown in Fig. 121 is bolted to the floor by six bolts
as shown. The pressure (or the pull) between the base of the crane and the
floor is assumed to be concentrated along the axes of the bolts. Find the ten-
PROBLEMS IN EQUILIBRIUM
117
sion in the two bolts to the left of the YY axis and the pressure between tEe
base and the floor at the corresponding bolts to the right of the axis, caused
by the load of 2000 lb. shown.
Ans. 2170 (tension); 2830 (compression).
Fig. 121.
102. Find R, the maximum unbalanced pressure (recoil pressure) that can
exist on the breech of a 3-in. gun (Fig. 123), when firing, without causing the
wheels to leave the ground, assuming the earth pressures and the weight of the
gun to be as shown in Fig. 123.
Fig. 123.
103. A uniform beam weighing W lb. rests with its ends on two smooth
planes which are inclined at angles of 60° and 30° with the horizontal. Find
the inclination of the beam with the horizontal. Ans. 0 = 30".
118
EQUILIBRIUM OF FORCE SYSTEMS
104. The two bodies shown in Fig. 124 are held in equilibri\im on two
smooth rods by a connecting cord. If the bodies weigh 200 lb. and 50 lb.,
find the reactions of the rods, the tension in the cord, and the angle 6.
106. The power press shown in Fig. 122 has the following dimensions:
AB = 2.5 ft. BC = BD = 1 ft. DE = 1.5 ft. FC = S ft. FB = 7.1 ft. The
points B, E, and F are on the same vertical line. What force P is required to
cause a pressure of 2000 lb. between the jaws of the press?
2001b.
Fig. 124:
Fig. 125.
106. Forces act along the sides of the polygon shown in Fig. 125. The
forces are proportional to the lengths of the corresponding sides. If the areas
of the two loops are equal, show that the system of forces is in equilibrium.
107. A body weighing 80 lb. is suspended by two strings of lengths 5 ft.
and 12 ft., the upper ends of which are attached to a horizontal plane, the dis-
tance between the two points of attachment being 13 ft. Find the tensions in
the strings. Ans. 30.8 1b; 73.8 1b.
108. A beam 8 ft. long rests on two horizontal supports and is loaded as
shown in Fig. 126. If the weights of the beams are neglected, find the reac-
tions Ri and R2 at the points of support.
^
E
8000 lb.
^=^=^
I I T I I I
Ri
FiQ. 126.
s
1000 lb.
Fig. 127.
PROBLEMS IN EQUILIBRIUM
119
109. A horizontal beam 10 ft. long is supported at its ends and loa^tod^
with two concentrated loads and a uniformly distributed load of 100 lb. per
ft. over a length of 4 ft. as shown in Fig. 127. Find the reactions Ri and Ri&t
the points of supports. Ans. Ri =980 lb. i?2 = 920 lb.
110. A camp stool loaded as shown in Fig. 128 rests on a smooth floor.
Find the reactions of the floor on the stool.
111. Find the magnitudes of the forces which must act along lines I and II
(Fig. 130) in order to hold in equilibrium the three forces shown. Solve by
the graphical method.
/SO lb.
ZiJ
Fig. 128.
Fig. 129.
112. In Fig. 132 is shown one form of a dynamometer. The pressure on
the scale beam at D is balanced by the poise weights A and B. The weight of
A and B together is 150 lb. and that of B alone is 3.5 lb. The divisions on
the large scale are 1 in. apart and those on the small scale are 0.4 in. apart.
II
201b.
I 10 lb.
301b.
Fig. 130.
Ill
"1^—30^ 2'U— 30^ 2'}^20^'
40-^^ s4-e 50-
^^
2^^
Fig. 131.
When both A and B are set at zero on the scales they are just balanced by the
weight C. If, in order to maintain balance, A is set at the 2nd division and B
120
EQUILIBRIUM OF FORCE SYSTEMS
is set at the 15th division, what is the pressure of the dynamometer on the
scale beam at D? Ans. i? = 26,75 lb.
113. In Fig. 131 is shown a diagrammatic sketch of an apparatus for
measuring a load on a beam. If a force of 5 lb. at £' is required to balance the
scale beam CDE when the beam AB carries a load P in the position shown,
find the magnitude of the load.
114. In Fig. 129 is represented a differential chain hoist. Two sheaves
of radii n and r2 are fastened together and a continuous chain passes around
the small sheave, then around a movable pulley of diameter n + ^2, and then
around the larger sheave. Neglecting the resistance due to friction, find
the relation between the applied force F and the load W which it will hold.
y
Fig. 132.
#
Hun
115. The bar shown in Fig. 134 is connected to a fixed support by a smooth
pin at A and rests on a smooth surface at B. Find the reaction at B and the
magnitude and direction of the pin pressure at A .
Ans. A =25 A lb.
116. A bar rests against two smooth surfaces as shown in Fig. 133 and is
prevented from slipping by means of a cord attached to the lower end. If the
weight of the bar is neglected, find the reaction of each surface and the tension
in the cord.
Ans. Ra =200 \h. /?s = 54.61b. r = 54.61b.
50 1b. „
80 1b.
^ I 50 lb.
m
Fig. 134.
PROBLEMS IN EQUILIBRIUM
121
117. In the crane represented in Fig. 135 the weight of the post AB in Fton,
the weight of the boom CE is 0.8 ton and that of the brace DF is 1 ton The
reaction at B is horizontal. Find the reaction at B and the horizontal and
vertical components of the reaction at A. Also find the pin reaction on the
brace at F and at />. Ans. ^ = 5.84 tons; Z> = 11.2tons; F = 11.9tons.
118. In the crane represented in Fig. 136, the weight of the post AB is
1600 lb., the weight of the boom DE is 1200 lb., and the weight of the rod CE
may be neglected. Find the tension in the rod and the reaction of the pin at
D on the post.
6 tons
Fig. 135.
Fig 136.
119. Two members AB and BC (Fig. 137) are connected by a smooth
pin at B and their lower ends rest on a smooth horizontal surface, slipping
on the surface being prevented by a cord which connects the ends A and C.
The weight of ^B is 120 lb. and the weight of BC is 180 lb. The member BC
Fig. 137.
122
EQUILIBRIUM OF FORCE SYSTEMS
also carries a concentrated load of 450 lb, as shown. Find the reactions of
the surface, the tension in the cord, and the pin reaction at B.
Ans. RA=^12\h. i?(7 = 378 1b. 7^ = 234 lb. /?B = 344 1b.
120. In Fig. 139 is shown a shear for cutting steel bars in a repair yard.
What force P is required to give a pressure of 70,000 lb. on the anvil when the
3 ft.-6 in. arm is vertical. Solve graphically.
121. Two bars AC and BD (Fig. 138) are connected to each other by a
smooth pin at C and to the floor by smooth pins at A and B. AC = CB = 4 ft.
CD = 3 ft. Find the pin pressures at A, B, and C.
Ans. A = 140 lb. B = 109 .8 lb. C = 1401b.
Fig. 138.
Fig. 139.
^.my/^:.
122. In Fig. 140, when r is zero, the tension in the spring, S, is 50 lb.
modulus of the spring is 58
lb. per inch. Find the ten-
sion in the spring in pounds
when r = 3 in. and also find
the pulls P, P required to
cause this spring tension.
What is the total stretch of
the spring?
Ans.^ = 94.1 lb. P = 24.3 lb.
Stretch = 1.62 in.
IP
The
500 lb. 1
Fig. 140.
Fig. 141.
PROBLEMS IN EQUILIBRIUM
123
123. The weights of the members AD, BC, and BD of the frame showu m
Fig. 141 are 100 lb., 50 lb., and 75 lb , respectively. Find the pin pressure at C
and the pressure at A, the surface at A being smooth.
Ans. ^ = 152 lb. C = 2251b.
124. Find by the method of sections, the stresses in the members CE, DE,
and DF of the Howe truss shown in Fig. 142.
2400 lb.
2000 lb.
125. The truss shown in Fig. 143 is supported at A and E. Find the
stresses in the members BD, CD, and CE.
126. Find the stresses in the members CE, DE, and DF of the truss shown
in Fig 144. Ans. CE=+UO lb.; D^= +500 lb.; DF= -550 lb.
2000 1b,
1000 1b.
10
00 1b.
1000 lb.
1000 lb.
y
r
1000 1b.
^
'^
''"^^-V^
.t
>
f^'
\
/
' !?
^
^
\
/
^\^
1
^ >
Panels @ 10'= (
^n'
Fig. 144.
Fig. 145
127. Determine graphically the stresses in the members of the Pratt truss
shown in Fig. 145.
128. The members of the crane shown in Fig. 146 are connected by smooth
pins. Neglecting the weights of the members, find by the graphical method,
the stresses in the members due to the 4-ton load.
124
EQUILIBRIUM OF FORCE SYSTEMS
129. The framework shown in Fig. 147 is connected at the points. A, D,
C, and E by smooth pins and is supported by a smooth surface at B. Find
Fig. 146.
the pin reactions at vl, D, C, and E, and the reaction at B. Neglect the weights
of the members.
1000 lb. ^ 1—1
Fig. 147.
130. In the hydrauHc crane shown in Fig. 148 the weight of the boom
ABC is 500 lb., as shown in the figure, and the weights of the members AD and
BD may be neglected. In raising the boom the hydraulic pressure from the
cylinder transmits a pressure P to the pin at D. Neglecting the frictional re-
sistance find the value of P required to raise the boom when loaded as shown.
PROBLEMS IN EQUILIBRIUM
125
Find also the stresses in BD and AD and the pressures of the rolls at JTaiid
at D against the vertical post.
Ans. F = 2500 1b.; D = A=5540 1b. Z)5 = +6550 lb. ; AD=-'im\h.
2000 1b.
131. The crank-pin pressures, Pi and Pz, on the crank shaft shown in Fig
150 are 6000 lb. and 4800 lb. respectively. Find the bearing reactions Ri
Fig. 149.
and R2 (Fig. 1505) and the resisting moment Qq required for equilibrium of
the shaft.
Ans. Ri = 3120 lb. R^ = 1920 lb. Qq = 64,800 lb .-in.
132. Three rods each of length / rest on each other (Fig. 149) in such a
126
EQUILIBRIUM OF FORCE SYSTEMS
way as to form a central equilateral triangle each side of which is f Z. A load P
is applied at one corner as shown. Find the reactions Ri, R2, and R3 in terms
of P.
Fig. 150.
CHAPTER IV
FRICTION
64. Friction Defined. — Friction is defined as the resistance
which one body offers to the motion of a second body when the
second body sHdes or tends to shde over the former. The fric-
tional force is tangent to the surfaces of contact of the two bodies
and always opposes motion.
Friction is of great importance in engineering practice. Since
it always opposes motion, it is an undesirable and expensive factor
in the operation of many machines or machine elements, and in
such cases is reduced as much as practicable by means of lubricants.
In other machines it becomes a very desirable and useful element,
as in various forms of brakes, friction drives, etc. In fact, many
of our normal physical activities, such as walking, would be impos-
sible without the aid of friction.
If the resistance between two bodies prevents motion of one
body relative to the other, the resistance is called static friction,
while the frictional resistance be-
tween two bodies which move rela-
tive to each other is called kinetic
friction. If the friction is static,
the amount of friction developed is
just sufficient to maintain equilib- "
rium with the other forces acting
on the body. That is, static friction
is an adjustable force the magni-
tude of which is determined from
the equations of equilibrium for the forces which act on the
body. Thus, let Fig. 151 represent a body in equiUbrium on
a rough horizontal plane under the action of a horizontal force
P, which tends to move the body, the reaction, R, of the
plane, and the weight, W, of the body. Let the reaction R be
resolved into two components F and N parallel and perpendicular,
127
W
sr
\r
\
Fig. 151.
128 FRICTION
respectively, to the plane. The component F along, or tangent to,
the plane is the frictional force as defined above. The component
N is called the normal pressure and R is called the total reaction.
Since the body is in equilibrium, the equations of equilibrium must
be satisfied, and hence F is equal to P. If the force P is gradually
increased, F must increase in the same ratio in order to maintain a
condition of equilibrium. There is a definite limit, however, to
the amount of frictional resistance that can be developed, and
when the value of P exceeds this limiting value motion will ensue.
The limiting or maximum value of the frictional force is called
limiting friction and is denoted by F'. Its value depends on
the normal pressure and on the roughness of the surfaces of
contact.
65. Coefficient of Friction. — In order to compare the frictional
resisting properties of various pairs of materials or of the same pair
of materials under varying conditions of their surfaces of contact,
and in order to calculate the maximum frictional force corre-
sponding to any normal pressure, a certain experimental constant,
called the coefficient of friction, is used.
The coefficient of static friction for any two surfaces is defined
as the ratio of the limiting friction to the corresponding normal
pressure. Thus, if the coefficient of static friction is denoted by
IX, it maj^ be expressed as follows :
F'
|JL= or F' = V'N.
N
It is important to note that F' in the above equation is the maxi-
mum friction which the surfaces can develop, that is, the
friction corresponding to impending motion. Thus the maxi-
mum frictional force which any two surfaces can develop is equal
to ixN.
The value of /x must be determined experimentally and, as
stated above, it is a constant for any two materials for a definite
condition of the surfaces of contact. It varies considerably,
however, for different conditions of the surfaces, and it varies
widely for different pairs of materials, as is shown in the following
table which gives the values of the coefficient of friction for dry
surfaces as determined by Morin and others.
ANGLE OF FRICTION 129
Coefficient of Static Friction
Wood on wood 0.25 to 0.50
Metal on wood 20 to .60
Metal on metal 15 to .30
Metal on leather 30 to .60
Wood on leather . 25 to .50
Stone on stone 40 to .65
Metal on stone 30 to .70
Earth on earth 25 to 1 .00
If two surfaces move relative to each other, the ratio of the
friction developed to the corresponding normal pressure is defined
as the coefficient of kinetic friction. The value of the coefficient of
kinetic friction for two surfaces is influenced by more factors than
is the value of the coefficient of static friction. A brief discussion
of the influencing factors is given in Art. 68. For values of the
coefficient of kinetic friction for various conditions of rubbing
surfaces the reader is referred to Good-
man's " Mechanics Applied to En-
gineering."
66. Angle of Friction and the Fric-
tion Cone. — The angle of static friction
t
for two surfaces is defined as the angle ' \
^
\
between the directions of the total
reaction and the normal pressure when
motion is impending. Thus in Fig. -p j^o
152, if the force P is just large enough
to develop the Hmiting friction, the angle which R, the reaction
of the plane on the body, makes with the normal pressure, iV,
is the angle of static friction and is denoted by 0.
Since the components of R, parallel and perpendicular respect-
ively to the plane, are F' and N, it is evident from the figure that
F' . . F'
tan <t> = -rr^. But since the ratio -^ is defined as the coefficient
N N
of static friction, /z, the following important relation may be
written :
\L= tan <j),
that is, the coefficient of static friction is equal to the tangent
of the angle of static friction.
If the two surfaces move relative to each other then the angle
between the total reaction and the normal pressure is called the
130
FRICTION
angle of kinetic friction. Its value is somewhat less than the angle
of static friction, since the frictional force after motion ensues
becomes less than the limiting friction. The relation /i = tan 0
also holds for kinetic friction, the value of n for kinetic friction
being somewhat less than for static friction. The angle of friction
(for both static and kinetic friction) is convenient to use particu-
larly in the solution of problems by graphical methods.
If a body rests on a rough plane and certain forces are applied
to the body, a convenient graphical method
of determining whether the body will sHde
or not makes use of the cone of friction.
The cone of friction for two plane surfaces is
the cone generated by revolving, about the
normal to the plane, a line making with
the normal an angle equal to the angle of
friction as shown in Fig. 153. If the re-
sultant of all of the forces acting on the
body, except the reaction of the plane (total
reaction), falls within the cone of friction,
the body will not shde; if the resultant falls
outside the cone of friction, the body will
sUde. Thus in Fig. 153 let P represent the
resultant of all of the forces acting on the
body except R, the reaction of the plane.
If the angle which P makes with the
normal to the plane is caUed d, the force tending to move the body
is P sin 6, and the normal pressure is P cos 6. The body will not
slide if
Psin e<iiP cos 0,
that is, if tan 6<fjL or if tan 6 < tan 0,
Fig. 153.
or if
0<0.
li d>(f), that is, if P falls outside of the cone of friction the body
will sUde, and if ^ = 0 motion is impending.
ILLUSTRATIVE PROBLEM
133. A lift shown in Fig. 154 slides on a vertical shaft having a square
cross-section 2 in. on a side. Find the greatest distance, x, that a load W
can be placed from the edge of the shaft and still cause the lift to slide on the
shaft. Neglect the weight of the lift and use 0.2 for the coefficient of friction.
CONE OF FRICTION
131
Graphical Solution. — Since motion impends, the angle between the reac-
tion Ri and the normal pressure A'^i of the shaft at A is equal to the angle of
friction; that is, the tangent of the angle is equal to 0.2. Hence, by laying
off ten spaces along the normal and two spaces perpendicular to the normal,
as shown in Fig. 154, the action line of Ri is determined. In a similar manner
the action line of R2 is found. Now the three forces Ri, R2, and W must be
concurrent in order to be in equilibrium (Art. 48) and hence, W must pass
through the intersection of Ri and R2. Now the intersection of these two
forces can never be nearer to the shaft than the point D, since the angle <l>
-> —
^3'H
1
E
Ni
r-0=tan-^ 0.2
H
, 1 , ., r
>
l^^~-h^^
^
F/
Ri
5
c
^
2"
f
Ro
0 = tan -1 0.2 ^
1
1
h
^^
J
r. __
B
^
\
^r , , , 1^1 1^1 >
^N.
^ «."
~^
^
wi
1
Fig. 154.
cannot be greater than tan-^2. The distance, x, of D from the shaft is found
by measurement to be 24 in.
Algebraic Solution. — The five forces, F\, Ni, F\, N2, and W which hold the
lift in equilibrium as shown in the free-body diagram (Fig. 154) form a coplanar
non-concurrent force system and hence there are three equations of equi-
librium as follows:
XFx=N2-Ni=0 (1)
W+0.2Ni-\-0.2N2=0 (2)
... (3)
in which
XMb= -Wx+10Ni-2X0.2Ni=0, .
0.2A^i = F'iand0.2iV2 = F'2.
From (1) and (2) we have,
W = 0.4Ni=0.4J^2.
By substituting this value of W in (3) the equation obtained is,
-0.4iVi-a:+10Ari-0.4 Ni = 0,
x=24in.
whence,
132
FRICTION
PROBLEMS
134. If the weight W in the preceding problem is 500 lb. what vertical
force P applied at E will be just sufficient to start the lift up?
135. The coefficient of friction for the 100-lb. body and the table (Fig. 155)
is 0.4. What weight, W2, will cause motion of the three bodies to impend?
The pulleys are assumed to be frictionless and the cords to be flexible.
Ans. Tr2 = 471b.
W3=10 lb. p
i2- >^
Fig. 155.
Fig. 156.
136. The ladder shown in Fig. 156 weighs 50 lb. The coefficient of
friction for the ladder and vertical wall is 0.25, and for the floor and ladder
0.5. Find the horizontal force P which will cause motion to impend to the
right. Ans. P = 5l lb.
-2 ft.
U
Fig. 158.
137. A rectangular block of wood 10 in. by 10 in. by 20 in. stands on end
on a horizontal floor. The block is acted on by a horizontal force applied
through the center of the top face and perpendicular to one side. The block
ANGLE OF REPOSE
133
weighs 40 lb. and the coefficient of friction is 0.20 If the force is gradually
increased until motion ensues will the block slide or tip? What is the mag-
nitude of the force when the block starts to move?
138. A crown friction drive as indicated in Fig. 157 is used on screw
power presses, motor trucks, etc. The cast iron disk B rotates at 1000
r.p.m. and drives the crown wheel C, which is faced with leather-fiber.
The diameter of C is 20 in. and the value of n is 0.35. If a turning moment of
100 ft. -lb. is transmitted to the crown wheel shaft when slipping impends, what
is the normal pressure between the disk and the crown wheel? What is the
pressure on the bearings at A and D?
139. A load, P, of 80 lb. is applied to the arm (Fig. 158) at a distance of 2
ft. from the axis of the sha£t. If ^i = 0.15 will the arm slide on the shaft?
140. A cone clutch as shown in Fig. 159 is used on automobiles to connect
the motor with the transmission. If the normal pressure between the two
Fig. 159.
surfaces of contact, as maintained by a spring, is 15 lb. per square inch and
the coefficient of friction is 0.25, what is the maximum torque the clutch
can transmit? Assume that the frictional force has a mean arm of 14^ in.
What spring pressure, P, is required to produce the normal pressure of 15 lb.
per square inch? Ans. T = 1235 Ib.-f t . ; P = 887 lb.
67. Angle of Repose. — If a body rests on an inclined plane,
as shown in Fig. 160, and is acted on by no forces except its weight
and the reaction of the plane, and if «, the angle of inclination of
the plane to the horizontal, is such that motion of the body
134 FRICTION
impends down the plane, the angle a is defined as the angle of
repose.
Since the body is in equilibrium under the action of the two
forces R, the reaction of the plane, and W, its weight, these forces
must be equal, opposite, and coUinear.
Hence the reaction R is vertical. Further-
more, the angle which R makes with the
normal to the plane is 0, the angle of fric-
tion. It is evident from the figure that the
angles a and </> are equal. The angle of
repose for two surfaces can be found eas-
ily by experiment, after which the coeffi-
cient friction for the surfaces may be found from the relation
/x = tan 0 = tan a.
68. The Laws of Friction. — One of the earliest contributions
to our knowledge of the laws of friction was made by Coulomb,
who pubhshed, in 1781, the results of experiments on the friction of
plane dry surfaces. Later experiments by Morin confirmed, in
the main, the results obtained by Coulomb. The results of the
experiments of Morin on dry surfaces, pubfished in 1831, may be
stated as follows :
1. The friction between two bodies when motion is impending
(Hmiting friction) is proportional to the normal pressure; that is,
the coefficient of friction is independent of the normal pressure.
2. The coefficient of static friction is independent of the area
of contact.
3. The coefficient of kinetic friction is less than the coefficient
of static friction and is independent of the relative velocity of
the rubbing surfaces.
Although these laws are probably correct for the conditions
under which the tests were made, they must be modified in order
to apply to friction which is developed under conditions quite dif-
ferent from those found in the experiments. The pressures used
in the experiments of Morin varied from J lb. per square inch to
100 lb. per square inch. It has been found in later experiments
that for pressures less than | lb. per square inch the value of the
coefficient of static friction increases somewhat. For very great
pressures the coefficient also increases. The highest velocity
used in Morin's experiments was 10 ft. per sec. For greater
velocities than this it has been found in later experiments that
THE INCLINED PLANE 135
the coefficient of kinetic friction decreases with the velocity.
The experiments of Jenkin show that for extremely low velocities
(the lowest velocity measured was .0002 ft. per sec.) there is an
increase in the coefficient of kinetic friction. These experiments
indicate that the value of the coefficient of kinetic friction grad-
ually increases as the velocity decreases and passes without dis-
continuity into that of static friction.
From experiments made by Tower, Goodman, Thurston, and
others, on lubricated surfaces, it has been found that the laws of
friction for lubricated surfaces are almost the reverse of those
stated for dry surfaces. For example, it is found that the friction
of two surfaces is almost independent of the nature of the surfaces
and of the normal pressure so long as there is a ffim of lubricant
between the surfaces. Again, for lubricated surfaces, it is found
that the friction is materially affected by the temperature, which is
not true in the case of dry surfaces.
69. The Inclined Plane. — Applications of the inclined plane
are found in various machines such as the jack-screw, the screw
type of testing machine, wedges, etc. Three variations of the
problem which arise in connection with the use of the inclined
plane will now be considered.
Case I. To Determine the Force Required to Start a Body up the
Plane. — Let Fig. 161 represent
a body resting on a plane, the
angle of incUnation of which
is a. It is required to find the
value of a force, P, which will
make motion of the body im-
pend up the plane when the
action line of P makes an
angle of 6 with the plane.
Although motion of the body
impends, it will not occur unless the value of P is increased
sHghtly beyond the critical value here considered. And since the
body is still in equilibrium the equations of equihbrium may be
applied to the system of forces acting on it. In addition to the
required force P, the forces acting on the body are the weight, W,
of the body, and the reaction of the plane. The latter force
will be resolved into components N, perpendicular to the plane,
and F' (equal to uN), parallel to the plane, as shown in the
Fig. 161.
136 FRICTION
free-body diagram. If the a;-axis be taken parallel to the plane
and the y-axis perpendicular to the plane, the equations of equi-
librium may be written as follows :
XFx=P cos d-fxN-W sin a = 0,
1:F„ = N+P sin ^-TFcosa = 0.
Eliminating N from the two equations we have:
p_TF (sin a+M cos a)
cos ^+M sin 6
If tan 4> be substituted for ijl this may be written,
p-j;^/Sin (a+0)^
cos {6—4))'
If the values of TF, a, and (/> are specified, P may be regarded as
a function of 0. The value of P is a minimum when 6 is equal to
0, since this value of 6 makes cos (^— 0) a maximum; the minimum
value of P, then, is W sin («+(/)). If the force P is apphed paral-
lel to the plane its value becomes W ; — - .
cos 0
Case II. To Determine the Force Required to Prevent Motion
Down the Plane when a > 0. — In this case the forces acting on the
body are the same as shown in Fig. 161 except that the frictional
force is directed up the plane. By writing the equations of
equilibrium as before and solving, the equation obtained is,
p_W sin (g — 0)
~ cos {d+<f)) '
For this case P is a minimum when 6= —<t), and the minimum value
of P is W sin («—</>). If the force P is applied parallel to the plane
the value of P is ^ ^.
cos 0
Case III. To Determine the Force Required to Start the Body
Down the Plane when a<cf). — In this case the forces which act on
the body are as shown in Fig. 161, except that P and jiN are
reversed. By writing the equilibrium equations as before and
solving, the equation obtained is,
p_TF sin ((t>—a)
cos (^+0) '
THE INCLINED PLANE
137
For this case P is a minimum when ^ = — </> and the minimum value
oi P is W sin {(t>—a). UP is appUed parallel to the plane, the
W sin i<j>—a)
value of P becomes
cos <f)
ILLUSTRATIVE PROBLEM
141. A small body weighing 50 lb. is placed on a rough plane which is in-
clined 30° with the horizontal (Fig. 162). The body is acted on by a force P,
the action line of which lies in the plane and makes an angle of 30° with the
line of greatest ^ope in the plane. If the coefficient of friction is ^, find the
value of P that will just start the block in motion, and find the direction in
which the block will begin to move.
Solution. — There are three forces acting on the body, namely, the force P,
the earth pull of 50 lb., and the reaction of the plane. The 50-lb. force may be
resolved into two components; 50 sin 30° in the plane (Fig. 162) and 50 cos
30° (not shown) perpendicular to the plane. The reaction of the plane may
be resolved into two components, N (not shown) perpendicular to the plane
and ^ N lying in the plane. The frictional force I N acts opposite to the
direction in which the body will begin to move. Let d be the angle which the
frictional force makes with the line of greatest slope. The two forces which
are perpendicular to the plane are in equilibrium and hence are equal. Thus,
Ar = 50cos30° (1)
The three forces in the plane are in equilibrium; hence, by Lami's theorem
(Art. 47),
50 sin 30° ^ P ^ ^N ,^.
sin (0+30°) sin d sin 30° ^ ^
By eliminating N from (1) and (2) the equation obtained is,
3 sin2 30° V3
2
Hence,
Therefore,
sm (0+30°)="
0+30'
cos 30
60°.
0 = 30"
Substituting 0 in (2) we have.
Fia. 162.
P=-^ = 14.4 lb.
2V3
PROBLEMS
142. A body weighing W lb. rests on a rough plane inclined at an angle 0
to the horizontal. What horizontal force must be applied in order to start
the body up the plane if the angle of friction is 0? Express the force in terms
<f> and also in terms of /x.
Ans. P = Trtan (d+<l>) = W
sin 0+M cos 0.
cos d—fjt sin 0
138
FRICTION
P- 120 lb.
143. A homogeneous rectangular prism, the dimensions of which are 1 ft.
by 1 ft. by 2 ft., stands on end on a flat square
board, the edges of the base of the prism being
parallel to the edges of the board. The coefficient
of friction for the board and prism is 0.2. If one
side of the board is gradually raised will the prism
slide or tip?
144. A body weighing 100 lb. rests on a rough
Fig. 163. plane inclined 45° to the horizontal. A horizontal
force of 50 lb. is just sufficient to prevent the body
from moving down the plane. Find
the coefficient of friction for the body
and plane. Ans. ai = 3-
146. Two bodies weighing 50 lb.
and 100 lb. rest on an inclined plane
and are connected by a cord which is
parallel to the line of greatest slope.
The body weighing 50 lb. is below
the one weighing 100 lb., and the
^efficient of friction for the 50-lb.
body is ^ and that for the 100-lb. body-
is J. Find the inclination of the plane
to the horizontal and the tension in
the cord, when motion impends.
146. A body weighing 200 lb. rests
on a plane inclined 30° to the hori-
zontal and is acted on by a force of
120 lb. as shown in Fig. 163. The
coefficient of friction for the body and
plane is 0 .3 . Find the friction between
the body and plane. Ans. 3.92 lb.
70. The Wedge. — Wedges are
used for lifting-devices, cotter-
pins, keys, etc. In the wedge
shown in Fig. 164(a), it is required
to find the value of the force P
which must be appHed to the
wedge to overcome the forces
W, W, appHed to blocks A and B.
It will be assumed that the angle
of friction for all rubbing surfaces
is the same. A free-body diagram
of the block A and of the wedge
Fig. 164.
THE WEDGE 139
C is shown in Fig. 164(6) and 164 (rf), respectively. The force P
may be found graphically by constructing the force polygons for
the forces acting on A and C. Since the force W is known com-
pletely and since the directions of the reactions {Ri and R2) of D
and C on A are known in direction, the force polygon for A can
be drawn as shown in Fig. 164(c). Having found the reaction R2,
it is now possible to complete the force polygon for the forces
acting on the wedge C. The reaction of ^ on C is equal and oppo-
site to the reaction of C on A, that is, it is equal to the force R2
just found, but reversed in sense. The force polygon for C is
shown in Fig. 164(e) from which P may be found. Fig. 164(/)
shows the two force polygons superimposed. From the force
polygons it is easy, by means of trigonometry, to derive the expres-
sion for P in terms of TF, </>, and a. It will be of interest, however,
to find the value of P by use of the equations of equilibriimi. Thus,
the equations of equiUbrium for the forces acting on A and C,
respectively, are,
XFx = W+Rismcf>-R2Cos(a+cf>) = 0, . . . . (1)
For A i
i:Fy = Ricos (f>-R2sm ia-\-<f>) = 0 (2)
^ ^ SFx=i^2 cos (a+</>)-ie3 cos («+</,) =0, .... (3)
For C \
i:Fy = R2sm(a+cf>)-\-Rssm{a+<ly)-P = 0. ... (4)
By eliminating Ri from (1) and (2) the equation obtained is,
_ TF cos (^
^^-COs(a + 20) ^^)
By ehminating R3 from (3) and (4) the equation obtained is,
P = 2R2 sin (a+<t>) (6)
By eliminating R2 from (5) and (6) the final equation is,
2W cos <f) sin (a-\-<f))
P =
cos {a-\-2(j>)
Another problem which arises in connection with the wedge is
the determination of the least force, P, which will prevent the
wedge from being forced upward by the loads W, W. The pro-
cedure in the solution of this problem is similar to that used in the
problem just discussed. The frictional forces, however, will be
reversed and hence the reactions will make angles 0 on the other
140
FRICTION
sides of the normals to the surfaces,
is given by the equation,
The value of P for this case
P =
2W cos (f> sin {a — 4))
cos {a — 24>)
When a is less than 4>, it is evident that the value of P is negative
and hence the wedge is self-locking, that is, it would stay in place
even if the force P were reduced to zero.
A third problem which arises in connection with the wedge is the
determination of the least pull, P, which is required to withdraw a
self-locking wedge, as for example, in lowering a heavy body by
means of the wedge. By writing the equations of equiUbrium and
solving, the value of the pull is found to be
p_2W cos 0 sin (0— «)
cos a
1000 lb.
ILLUSTRATIVE PROBLEM
147. Fig. 165(a) represents a cotter joint. The angle a equals 15° and the
angle of friction for all rubbing surfaces is
12°, What is the value of the force P re-
quired to overcome the 1000-lb. forces
applied on parts A and Bl
Solution. — Free-body diagrams of the
block A and the cotter pin C are shown
in Fig. 165(6) and 165(c) respectively.
The equations of equilibrium for the two
blocks may be written as follows:
For A
S/^a: = 1000 -f-/?i sin 12°
-7^2 cos 27° = 0, (1)
SFj/ = i2iCosl2°-722sin27° = 0. . (2)
ForC
XFx = R2 COS 27° -7^3 cos 12° =0, (3)
i:Fy = R2sin27°-}-R,sml2°-P = 0. (4)
By eliminating Ri from (1) and (2) the
equation obtained is,
R,
1000 cos 12'
cos 39°
(5)
THE WEDGE
141
By eliminating R3 from (3) and (4) the following equation is obtained:
P cos 12°
R2 =
(6)
sin 39°
By equating values of R2 in (5) and (6) the value of P may be found. Thus,
P cos 12° 1000 cos 12°
Therefore,
sin 39° cos 39° *
P = 1000tan39°=8101b.
PROBLEMS
148. The load of 100,000 lb. (Fig. 166) is raised by applying forces P, P
to the wedges. What is the required value of P if the coefficient of friction is
0.2 for all surfaces of contact? Ans. P = 26,667 lb.
Fig. 166.
149. Find the value of the force P which must be applied to the wedge A
(Fig. 167) in order to raise a weight W carried by the block B, assuming that
the coefficients of friction for all rubbing surfaces are equal.
Ans. P = TFtan (20+a).
150. Find the expression asked for in Problem 149 if friction is neglected
at all surfaces except that between the block B and the wedge A .
1000 lb.
^^^^^^^^^^^^^^^^^^
'm/////M
1000 lb.
m^/m/^
Fig. 167.
Fig. 168.
151. Find the force P which must be applied to the wedge shown in Fig.
168 in order to overcome the 1000-lb. forces applied to blocks A and B as
142
FRICTION
shown. Assume the coefficient of friction for all rubbing surfaces to be
0.1. Ans. 467 lb.
152. What is the value of P in Problem 151 if the horizontal surfaces, on
which blocks A and B rest, are smooth?
71. The Screw. — A screw is, in effect, an inclined plane wound
around a cylinder. Screws are made with square threads and
with triangular threads,
w but square - threaded
screws, only, will be con-
sidered here. Fig. 169(a)
shows a jack-screw with
square threads which is
used in raising or lower-
ing heavy loads. The
radius of the base of the
thread is denoted by ri
and the outer radius by
r2; a is called the pitch
angle and p is called the
pitch of the screw. Let
it be required to find the
force, P, which, when ap-
pUed at the end of a
lever of length a, is just
sufficient to raise the load
W. The forces which
hold the screw in equi-
librium are: the force P;
the pressure of the cap,
C, on the head, H; and
the reaction of the nut,
B, on the screw. The
latter is distributed over
the area of the threads
in contact with the nut.
If the friction between
Fig. 169.
the cap and the head of the screw is neglected, the pressure
of the cap on the head of the screw will be a vertical force equal
to W. The problem will first be solved on this assumption.
Two of the equilibrium equations which apply to this type of force
THE SCREW 143
system will be sufficient for the solution of the problem, namely,
2^2 = 0 and l^Mg = 0 where z is taken as the axis of the screw.
The reaction between the nut and the thread of the screw on an
element of area dA will be denoted by dR. This force may be
resolved into components dN normal to the thread and dF' par-
allel to the thread as shown in Fig. 169(6). In taking moments
about the axis of the thread it will be sufficiently accurate to con-
sider the moment-arm of dF' to be equal to the mean radius of the
thread, J (ri-|-r2), which will be denoted by r. The equiUbrium
equations stated above, then, become:
l:F^= XdN cos a-i:dF' sin a-W = 0,
XMz = Pa- i:rdN sin a - ZrdF' cos a = 0.
Since dF' = iidN, these equations may be written:
cosQ:2(iA^-MsinQ!2€?A^-TF = 0, (1)
Pa-r sin aSrfA/'- Mr cos a2diV = 0 (2)
By eliminating l^dN from (1) and (2) the equation obtained is,
T^ TT7- sin a -f At cos Of
cos a — yi sm a
By substituting tan </> for ju this equation may be written in the
form,
Pa = TFr tan (0+q:).
If the pitch angle a is large and the angle of friction is small,
the load W will cause the screw to run down unless a force is
appUed to prevent it. The force P required to hold the load
is found by a method of analysis similar to that used above, the
only difference being that the sense of the frictional force is
reversed. The least value of P required to prevent the screw from
running down is given by the equation,
Pa = TFr tan (q;-0).
If a=<f> in the above equation, the force P reduces to zero,
that is, the load will be held by friction alone. If a < 0, a force
144
FRICTION
is required (the sense of which is opposite to that in the two
cases considered) to lower the load. The value of the force
required to lower the load is given by the equation,
Pa = TFr tan (</)-«).
In practice, jack-screws are always made self -locking, that is, a is
made less than </>. The values of P for the three cases just con-
sidered will be changed somewhat if friction between the cap and
the head of the screw is considered. If the area of contact between
the cap and the head is a fuU circle the effective arm of the fric-
tional resistance is two-thirds the radius of the circle as will be
shown in Art. 74. In practice, however, the area of contact is
usually a hollow circle, and it wiU be sufficiently accurate to take
the mean radius as the effective arm of the frictional resistance.
If the mean radius is denoted by r' and the coefficient of friction
between the cap and the head of screw by ix, the values of P are
given by the following equations:
To raise the load, Pa=Wr tan {<f)-\-a)-^fxWr\
To hold the load. Pa = Wr tan {a-<f>)- txWr',
To lower the load, Pa = Wr tan {(f)—a)+uWr'.
PROBLEMS
163. The mean diameter of the screw of a square-threaded jack-screw is
1.8 in. The pitch of the thread is
0.4 in. and the coefficient of friction
for the screw and nut is 0.12. What
force must be applied at the end of a
lever 18 in. long to raise a weight of
5000 lb.? What force is required to
lower the weight?
Ans. 48.1 lb. 12.3 lb.
154. A weight of 1000 lb. is lifted
by applying a couple, Pd, to the
hand wheel of the apparatus
shown in Fig. 170 The diameter.
d, of the hand wheel is 20 in.; the
mean diameter of the screw is 1.5
in.; the pitch of the thread is i in.;
and the coefficient of friction is 0.15.
Find the values of the forces of the
couple when the value of ^ is 15°.
W=1000 lb.
Fig. 170.
JOURNAL FRICTION AND FRICTION CIRCLE
145
Fig. 171.
155. The shaft-straightening hand press shown in Fig. 171 is used" for
bending or straightening 3^ in. steel shafts. What force, P, applied at the
end of a 36-in. lever is required to
produce a pressure, Q, of 24,000 lb.
on the shaft? The threads have a
mean diameter of 2 in. and there
are four threads per inch. Con-
sider friction between the screw and
nut only, and use a value of 0.2 for
the coefficient of friction.
Ans. P = 161 lb.
72. Journal Friction and
the Friction Circle. — In Fig.
172 is shown an axle or jour-
nal in a bearing, the diameter
of which is slightly greater
than the diameter of the axle.
The difference in the diameters
is exaggerated in the figure
for the sake of clearness.
Since the contact between the axle and the bearing is along a line,
the reaction of the bearing on the axle is distributed along the
line of contact and may be replaced by a resultant force, i2, at
A, the midpoint of the line of contact.
If the axle rotates in the bearing or, if
motion impends, the angle between the
action line of the reaction, jR, and the
normal to the surfaces at A is the angle
of friction </>. In the former case the
angle is the angle of kinetic friction and
in the latter case it is the angle of static
friction. If the radius of the axle is
denoted by r, a circle having a center
which coincides with the center, 0, of
the axle and which has a radius equal to
r sin 4> will be tangent to the action line
of R. This circle is called the friction
circle. It will be noted that when the angle <i> is small, sin </> is
approximately equal to tan 4> and hence also approximately
equal to /x. Therefore, for small values of /x the radius of the
friction circle may be taken as /xr without serious error.
Fig. 172.
146
FRICTION
The friction circle is of importance in locating the line of action
of the reaction between an axle and its bearing since this line is
tangent to the friction circle when motion occurs or is impending.
For example, in Fig. 172 the hne of action of the reaction is one of
the two tangents that can be drawn from A to the friction circle.
The direction of motion (or impending motion) will determine
which one of the two tangents should be used. Thus, if the axle
rotates, or is about to rotate, in a clockwise direction the action
line of the reaction is as indicated in Fig. 172. In general the posi-
tion of the line of contact between the axle and bearing is not
Fig. 173.
initially known. This position can usually be determined, how-
ever, by means of the friction circle. For example, consider the
bell-crank, shown in Fig. 173(a), which is acted on by a known
force Q. Let it be required to determine the magnitude of the force,
P, which must be appHed along the line OB in order to cause
motion to impend in a clockwise direction. The forces acting
on the bell-crank are Q, P, and the reaction, R, of the bearing.
Since the three forces are in equiUbrium they must be concurrent.
Hence if P and Q intersect at 0, R must also pass through 0.
Furthermore, R must be tangent to the friction circle. There are
JOURNAL FRICTION AND FRICTION CIRCLE
147
only two tangents that can be drawn from 0 to the friction cifcIeT
The one to be used is indicated in the figure. Therefore, the loca-
tion of the line of contact between the axle and the bearing is at A.
Since the action Hne of R is now known, the magnitudes of P and R
can easily be found by drawing the force polygon (Fig. 1736) for
the three forces. If motion impends in the opposite direction and
it is required to find the least value of the force P which will
maintain equilibrium the other tangent would be used.
ILLUSTRATIVE PROBLEM
156. Fig. 174 represents a pulley, having a diameter of 2 ft., mounted on a
2-in. axle. The coefficient of friction between the axle and the bearing in
which it rests is 0.15. What is the least
value of the force P which will raise the
200-lb. weight, assuming that the fric-
tion between the rope and pulley is
sufficient to prevent slipping? Neglect
the weight of the puUey.
Solution. — ^There are three forces act-
ing on the pulley: the 200-lb. force, P,
and 22, the reaction of the bearing.
Since these three forces are in equi-
librium they must be parallel (Art. 48).
Hence R must be vertical and, since it
must also be tangent to the friction
circle, its line of action is determined.
The radius of the friction circle is nr or
Fig. 174. 0.15 in. The equations of equilibrium
are:
Sify = i2-P-200=0.
SMo = 200Xl2+0.15i2-12P = 0.
Eliminating R from these equations we have,
P = 2051b.
If the force R acted through the center of the axle, that is, if the axle were
smooth, the value of P would be 200 lb.
148
FRICTION
PROBLEM
157. A load of 1000 lb. is raised by a wooden wheel and axle as shown in
Fig. 175. What is the least value of the force P which will raise the load if the
coefficient of friction between the axle and the bearing is 0.4?
1000 lb.
Fia. 175.
Fig. 176.
73. Pivot Friction. — ^^In Fig. 176 is shown a pivot in a step
bearing. The pivot will be assumed to turn in its bearing and
hence kinetic friction, only, is involved. In order to determine
the resisting moment due to the frictional forces at the end of a
pivot, certain assimiptions will be made as follows:
1. The coefficient of friction is constant over the end of the
pivot.
2. The pressure between the pivot and the bearing at any point
(a) is constant, or
(6) varies in such a manner that the wear of the pivot in the
direction of its axis (axial wear) is uniform over the
area of the pivot. Or, as sometimes stated, the wear,
at any point, normal to the rubbing surfaces (normal
wear) is proportional to the work of friction.
Since the value of the coefficient of kinetic friction varies some-
what with the velocity of the rubbing surfaces, it is evident that
the coefficient is not constant over the end of the pivot, for the
velocity of any point varies as its distance from the axis of the
pivot. The assumption made in (1), however, will not cause
SOLID FLAT PIVOT 149
serious error if a mean value of the coefficient is used. For a new
pivot the assumption 2(a) seems reasonable, although the pressure,
probably, will not remain constant after wear occurs. If the wear
normal to the pivot (normal wear) at any point of the pivot is
proportional to the work done by the frictional force at that point,
that is, to the product of the frictional force, F', and the distance
traveled by the point, the normal wear will also be proportional
to the normal pressure, since F' is equal to fxN. If, then, the
normal pressure is constant, it is evident that, for a time, the
greater wear will occur near the circumference of the pivot.
The effect of this greater wear at the edge of the pivot will be
to increase the intensity of pressure near the center and to decrease
it near the edge of the pivot. The increased pressure near the
center will in turn result in increasing the wear near the center
and in decreasing it near the edge. This process will continue
until a uniform condition has been established, after which, the
wear will be uniform over the end of the pivot. The assumption
made in 2(6), therefore, seems reasonable for pivots that have
been in use for some time. The frictional moment will now be
determined for several pivots.
74. Solid Flat Pivot. Uniform Pressure. — In the flat pivot
shown in Fig. 176 the axial load will be denoted by TT, the radius
of the pivot by r, and the area of the pivot by A. The frictional
moment may be found as follows:
W
Pressure per unit of area = — g.
W W
Pressure on element of area = — :^dA= — r, pdpdd.
uW
Frictional force on element dA = — ^ pdpdO.
wr^
uW
Moment of frictional force on dA =^— o- p^dpdd.
^ jo i '
Total frictional moment = — ^ I I p^dpdB
=%pWr.
Uniform Wear. — Since the wear at any point is proportional to
the distance traveled by that point, it is proportional to the dis-
150
FRICTION
tance, p, of the point from the axis of the pivot. The wear, how-
ever, is also proportional to the friction force and hence, to the
pressure, p. That is, pp is equal to some constant, C. The
frictional moment, then, may be found as follows:
Frictional moment
-a:
Jo Jo
= )uC7rr2.
fjLpp^dpdd
pdpdd
(1)
But, the total pressure must be equal to W. Hence, another
equation containing C may be found. Thus,
W
= I I ppdpdd
Jo Jo
' j ^ j'dpdd
--C
= 2irCr
W
(2)
^^
I
I
I
I— r^
I
I
(a)
Fig. 177.
m^
By eliminating C from equations (1) and (2) the resulting expres-
sion is,
Frictional moment = \p,Wr.
75. Hollow Flat Pivot or Collar Bearing. — The determination of
the frictional moment for a hollow flat pivot (Fig. 177a), or collar
bearing (Fig. 1776), involves the same steps as were used in finding
CONICAL PIVOT
151
the f Fictional moment for a solid fiat pivot and, hence, will not be
given in detail. The f rictional moment is found to be :
2MTF(r2^-ri3)
For uniform pressure
For uniform wear .
'3 r22-ri2
yw{n+r2).
Uniform Pressure.-
76. Conical Pivot. Uniform Pressure. — In finding the fric-
tional moment for a conical pivot (Fig. 178) it will first be assumed
that the intensity of pressure, p, normal to the surface is constant.
The frictional force at any point of course is horizontal. Referring
to Fig. 178, we find the frictional moment as follows:
Frictional force on an element
of area dA = nydA ; W
Moment of this force
= uppdA ;
Total frictional moment
/
ixppdA
= I fjLpp2Trp
Jo sm a
_2Tr fxp r^
sin a 3 *
(1)
mmm
W///M//
Fig. 178.
The value of the constant pressure p in (1) may be expressed
in terms of W as follows : Since the sum of the vertical components
of the pressures on the elements of area must be equal to the axial
load TF, we may write,
=/
W = I pdA sin a = p sin
afdA =
pA sin a,
where A is the area of contact between the pivot and bearing.
And, since A Xsina^wr^ the above equation may be written,
W = pTrr^ (2)
By eliminating p from equations (1) and (2) the final expression is,
2 fiWr
Frictional Moment
3 sin a*
152
FRICTION
Uniform Wear. — If the wear in the direction of the axis of the
pivot is constant it can be shown that,
The frictional moment =
The proof is left to the student.
PROBLEMS
1 fj,Wr
2 sin a
168. The weight of the vertical shaft and the rotating parts of a turbine is
100,000 lb. and the diameter of the shaft is 10 in. Assuming the coefficient
of friction to be 0.015 and the bearing to be a flat-ended pivot, find the fric-
tional moment (1) when the pressure is uniform and (2) when the wear is
uniform.
169. Find the moment of the friction on a collar bearing, when subjected
to a pressure of 6000 lb., if the radii of the collar are 3.5 in. and 4.5 in. and the
coefficient of friction is 0,025, Assume the pressure to be uniform,
160. Derive an expression for the frictional moment for the spherical
pivot shown in Fig, 179 for each of the following assumptions: (1) uniform
normal pressure; (2) uniform wear in the axial direction.
Am. (1) W^rl"-'^. """"')■,
\ sm2 a I
(2)
W\xr sin^ a.
a+sin a. cos OL
Fig. 179.
Fig. 180.
161. Determme the frictional moment for the conical pivot shown in
Fig. 180 for each of the following assumptions: (1) uniform normal pressure;
(2) uniform axial wear.
4^. (1) 2^.'^^. (2)Tf^;^'+^'
3 sin a. n^—n^'
2 sin
ROLLING RESISTANCE
153
77. Rolling Resistance. — If a rigid wheel or roller which car-
ries a vertical load rests on a rigid horizontal surface, a horizontal
force, however small, will cause the wheel or roller to roll on the
surface. If a wheel rolls over a yielding surface, however, a
resistance to the motion is encountered due to the fact that the
surface immediately in front of the wheel is being deformed.
In Fig. 181 is shown a wheel carrying a vertical load W. Let
P be a horizontal force which causes the center of the wheel to
move with a constant velocity.
Since the surface on which the
wheel rolls deforms under the
wheel, the pressure between the
wheel and the surface is dis-
tributed over the area of con-
tact. The resultant pressure or
reaction of the surface on the
wheel, then, passes through some
point, B, in the area of contact as
shown in the figure. Since the
velocity of the wheel is constant,
the three forces acting on it are in equilibrium and hence the
reaction, R, of the surface on the wheel must pass through 0, the
center of the wheel. Taking moments about B we have,
Fig. 181.
i:Mb=wxab-pxoa=o.
Since the depression is usually small, OA is approximately equal
to r, the radius of the wheel. By using this approximation and
denoting AB by a, the value of P is found to be,
P =
Wa
The force P is equal to the horizontal component of the reaction R
and is called the rolling friction or rolling resistance; the distance a
is sometimes called the coefficient of rolling resistance. However,
since a is a hnear quantity and not a pure number it is not a true
coefficient. The value of a is generally expressed in inches. The
laws of roUing resistance are not well known, and there is need of
further investigation on the subject. It was assumed by Coulomb
I hat the coefficient of rolling resistance is independent of the radius
154
FRICTION
of the wheel. Tests by Dupuit indicate that the coefficient varies
as the square root of the diameter. Whether the conclusion of the
latter is correct or not, it seems reasonable to assume that the
value of the coefficient depends on the diameter of the wheel.
The values of the coefficient of rolling resistance given by various
investigators are not in close agreement and should be used with
caution.
Coefficients of Rolling Resistance
(Due to Coulomb and Goodman)
a (inches)
Lignum vitse on oak 0.0195
Elm on oak 0327
Steel on steel.. .' , 007 to 0.015
Steel on wood 06 to .10
Steel on macadam road 05 .20
Steel on soft ground . 3.0 to 5 . 0
Pneumatic tires on good road 0 . 02 to 0 . 022
Pneumatic tires on mud road 04 to .06
Solid rubber tire on good road 04
Solid rubber tire on mud road 09 to .11
The resistance due to rolling may be regarded as equivalent
to a couple. In order to show that this statement is true, con-
sider the wheel represented in Fig. 182. A flexible cord passing
over the wheel carries at its ends
weights W and W-\-w. Assume
w to be just great enough to
cause the wheel to move with a
constant velocity. Since the
wheel is in equilibrium under the
action of the two vertical forces,
W and W+w, and the reaction,
R, of the surface at B, it is evident
that the latter force must be ver-
tical. The reaction R at B may
be replaced by an equal parallel
force through A and a couple having a moment equal to Ra.
Since the force, R, through A also passes through the center of the
wheel it will not effect the rotation. Therefore the rolling resist-
ance is equivalent to the couple Ra. The moment wr, then,
which causes the wheel to roll is equal to Ra.
Fig. 182.
BELT FRICTION 155
For moving heavy weights, such as machines, houses, etc., two
or more rollers are frequently used as shown in Fig. 183. If the
coefficient of rolUng resistance is the same at the top and bottom
of the rollers it can be shown that the force P required to overcome
friction is given by the expression,
p^Wa
r
w
■mdm^mdmmdm^
Fig. 183.
PROBLEMS
162. The rolling resistance for the wheels of a freight car is 3 lb. per ton.
If the diameter of the car wheels is 33 in., what is the coefficient of rolling
resistance? Ans. 0 025 in.
163. An oak beam which carries a load of 5000 lb. rests on elm rollers the
diameters of which are 6 in. The rollers rest on a horizontal oak track. What
horizontal force is required to move the load if the weight of the beam is
neglected?
164. What is the rolling resistance of a wagon wheel on a macadam road
if the diameter of the wheel is 4 ft. 6 in.? Assume the coefficient of rolling
resistance to be 0.2 in.
Ans. 14.8 lb. per ton.
78. Belt Friction. — Belt friction is important in the trans-
mission of power by belt and rope drives and in resisting large
loads by means of band brakes, capstans, etc. If a belt, rope, or
steel band passes over a smooth cyhnder or pulley the tensions in
the belt, rope, or band on the two sides of the pulley are equal.
If the cylinder or pulley is rough, however, the tensions will not,
in general, be equal. In the present article the relation between
the tensions in the belt, etc., on the two sides of a rough pulley,
when the belt is about to slip, will be determined. It is evident that
the greater tension must be just 'large enough to overcome the
smaller tension in addition to the friction between the belt and the
pulley. In Fig. 184(a) is represented a belt on a pulley, the angle
of contact being a and the belt tensions being Ti and T2. Let
156
FRICTION
Ti be the greater tension and let it be assumed that the belt is
about to slip on the pulley. The normal pressure between the
belt and the pulley per unit length of belt, that is, the intensity of
pressure at any point, will be denoted by p, and the tension in the
belt at the same point will be denoted by T. Fig. 184(6) is a free-
body diagram of an element of belt of length ds. The forces
acting on this element are the tensions T and T-\-dT at the
ends, and the reaction of the pulley. The latter force may be
resolved into a component, dN = pds, normal to the face of the
T + dJ
Fig. 184.
pulley and a frictional component, dF' = iJipds, tangent to the face
of the pulley. The equations of equilibrium may be appUed as
follows :
XF,= {T+dT) cos ^-Tcos -^-fipds^O.
de
de
'LFy = pds-{T+dT) sin ^-T sin — = 0.
(1)
(2)
Since — is small, cos -7^ is approximately equal to unity and
sin — is approximately equal to -^. The term dT sin -- i;j a small
quantity of the second order and may be neglected. By using
these approximations, equations (1) and (2) become,
dT-txpds = 0,
pds-Tdd = 0,
(3)
(4)
BELT FRICTION
157
Eliminating pds from equations (3) and (4) we have,
(5)
By integrating equation (5) the relation between Ti and T2 may-
be found as follows:
That is,
or,
1 ^1
Ti
= e'^,
T2
(6)
where e is the base of natural logarithms and a is measured in
radians. It should be noted that in the derivation of equation
(6) the belt is assumed to be perfectly flexible.
ILLUSTRATIVE PROBLEM
165. In the band brake shown in Fig. 185 the force P is 100 lb., the angle
of contact, a, is 270° (Itt radians), and the coefficient of friction, ju, for the band
and the brake wheel is 0.2. If the brake wheel rotates in a counter-clockwise
direction find the tensions in the band and the frictional moment developed.
Fig. 185.
Solution. — Since the operating lever ACB is in equilibrium the equation
SAf B = 0 may be applied, from which the band pull at C, that is, the tension
T2, is found. Thus,
SMb = 100X26- To X2=0,
.-. 2^2 = 1300 lb.
158 FRICTION
Since Ti is now known, the tension Ti may be found from the belt-friction
formula. Thus,
7'i = T2Xe^
= 1300X(2.718)0-2x|^,
log T,=log 1300+0.37r log 2.718
= 3.114 + .942X.434
= 3.522.
/. Tt = 33301b.
Frictional moment = (Ti - T^) X 10
= 2030X10 = 20,300 Ib.-in.
PROBLEMS
166. A body weighing 2000 lb. is suspended by means of a rope wound
\\ turns around a drum. If the coefficient of friction is 0.3 what force must
be exerted at the other end of the rope to hold the body?
167. A boat is brought to rest by means of a rope which is wound around a
capstan. If a force of 4000 lb. is exerted by the
boat and a pull of 100 lb. is exerted on the other
j^gg^J end of the rope, find the number of turns the rope
makes around the capstan, assuming the value of
M to be 0.25. Ans. 2.36 turns.
168. A body weighing 500 lb. is raised by
means of a rope which passes over a roimd
beam, the angle of contact being 180°. If the
coefficient of friction is 0.4, what is the least force
which will raise the body? What is the least
force which will hold the body?
169. A rope is woimd twice around a post.
If a pull of 50 lb. at one end of the rope will just
support a force of 6000 lb. at the other end,
i/y/////////////^ what is the coefficient of friction?
Fig. 186. 170. A body having a weight, TF, of 1 ton is
suspended by means of a wire rope which passes
over two fixed drums, as shown in Fig. 186. If the coefficient of friction for
the rope and drums is 0.3, what force, P, will be required (a) to hold the
body; (6) to start the body upwards?
Am. (a) P=488 1b.
BELT FRICTION 159
171. In Fig. 187 is represented a band brake, the angle of contact onhe"
P-601b.
Fig. 187.
band on the brake wheel being 180°. If the coefficient of friction is 0.2, find
the frictional moment developed (a) when the brake wheel rotates clock-
wise; (h) when the brake wheel rotates counter-clockwise
Ans. (a) 157 Ib.-ft.
CHAPTER V
FIRST MOMENTS AND CENTROIDS
79. First Moments. — In the preceding chapters moments of
forces with respect to points and hnes have frequently been used.
In the analysis of many problems in engineering, however, expres-
sions are frequently met which represent moments of volumes,
masses, areas, and lines. The moment of a volume, mass,^ area,
or line with respect to an axis or plane is the algebraic sum of the
moments of the elementary parts of the volume, mass, area, or line,
the moment of an elementary part being the product of the ele-
mentary part (volume, mass, area, or Une) and its distance from
the moment axis or plane. This moment of a volume, mass, etc.,
is called the first moment when it is desired to distinguish it from
the moment of inertia (or second moment) of the volume, area,
etc. (see Chapter VI), since the coordinate distances of the parts
of the volume, mass, area, or Une enter into the expression for the
first moment to the first power and into the expression for the
second moment to the second power.
80. Centroids. — In dealing with a system of parallel forces in
a plane it was found from the principle of moments that the alge-
braic sum of the moments of the forces about a point or axis is
equal to the moment of the resultant of the forces with respect
to the same point or axis. That is, 'L(F'x) = ZF'X = Rx (Art. 26).
This equation was frequently used in Chapter II to locate the
position of the resultant, that is, to determine x after first finding
the sum of the moments of the forces, ZiF-x). When the posi-
tion of the resultant is known, however, the moment of the system
of forces is most easily found by determining the moment (Rx) of
the resultant.
^ The term mass cannot be defined completely nor its physical significance
discussed until the laws of motion of physical bodies are treated, (See
Chapter IX.) As here used it is sufficient to think of mass as the inert mate-
rial or matter of which bodies are composed, the quantitative expression of
which is the volume of the body times a density factor.
160
CENTROIDS 161
In a similar way the moment of a volume, mass, area, or Ene~
with respect to an axis or plane (made up of the sum of the moments
of the several parts or elements of the volume, mass, area, or line)
may be expressed as the product of the whole volume, mass, area,
or line, and a distance, x (or y, etc.), from the axis or plane such
that this product equals the algebraic sum of the moments of the
elementary parts of the volume, mass, area, or line.
This relation is sometimes called the principle of moments for
areas, volumes, etc., and it leads to an equation of the same form as
does the same principle in the case of forces. Thus, for an area A,
an element of which is denoted by a or dA, the equation is,
l^(ax) = 'Za-x '=Ax,
j(dA-x) = Jc
or
(dA •x)= \ dAx^Ax.
The resulting distance (x or y, etc.), is called the centroidal dis-
tance, and the point located by the centroidal distances is called
the centroid ^ of the volume, mass, area, or Hne. The centroid of a
volume, mass, area, or line, then, is that point at which the whole
volume, mass, area, or line may be conceived to be concentrated
and have the same moment with respect to an axis or plane
as has the volume, mass, area, or Hne when distributed in its
natural way.
Hence the coordinates {x and y) of the centroid of an area,
A, and of a line, L, are expressed by the following equations, in
which A = Sa= | dA and L = ^l= | dL.
^ The term center of gravity is sometimes used in technical Hterature to
denote not only the point in a body through which the resultant earth-pull
acts, but also to denote what is here defined as the centroid of a line, area,
volume, or mass. Thus, the phrase " center of gravity " of an area, or vol-
ume, etc., is used instead of the phrase " centroid " of the area, or volume, etc.
Further, the term centroid is sometimes used in a restricted sense as applying
only to geometrical figures (lines, areas, and volumes), in which case, the
term mass-center or center of mass is used instead of centroid of mass in con-
nection with physical bodies.
162 FIRST MOMENTS AND CENTROIDS
^ A A ' y~ L Z~'
Likewise, the coordinates (x, y, and ~z) of the centroid of a volume,
V, and of a mass, M, may be expressed by similar equations, in
which V=i:v= I dV and ilf =2m= j dM. Thus,
/xdV ^r \ I ■
-_^(mx)_J
xc^M
__2:M
/l/c^F
7 7 '
M
M '
X{my)
M
^ydM
M '
Umz)
CzdM
7 7' ikf M •
81. Planes and Lines of Symmetry. — If a geometrical figure
(volume, area, or line) is symmetrical with respect to a plane or
Une, the centroid of the figure lies in the given plane or line.
This statement is evident from the fact that the moments of the
parts of the figure on the opposite sides of the plane or line are
numerically equal but of opposite sign. If a figure is symmetrical
with respect to each of two planes or lines, the centroid of the
figure lies in the line of intersection of the two planes or at the point
of intersection of the two fines. If the figure has three planes of
symmetry, the centroid coincides with the point of intersection of
the three planes. The foregoing statements apply also to the
centroids of the masses of homogeneous physical solids which are
symmetrical with respect to one or more planes, since the centroid
of a volume coincides with the centroid of the mass of a homo-
geneous body which is congruent with the volume. The centroids
of many simple figures may be partially or completely determined
from symmetry. Thus, the centroids of the volumes or of the
surface areas of the following solids are as indicated below:
(1) Sphere or ellipsoid; the center of the sphere or elHpsoid.
(2) Right prism or cylinder; the mid-point of the axis.
(3) Hemisphere; on the radius perpendicular to its base.
(4) Right cone; on its geometrical axis.
CENTROIDS BY INTEGRATION 163
And the centroids of the following areas, and of their bounding
lines, are as indicated below:
(5) Circle or ellipse; the center of the circle or eUipse.
(6) Isosceles triangle; on the median bisecting the angle
between the equal sides.
(7) Semicircle; on the radius perpendicular to the base.
82. Centroids by Integration. — In determining the centroid of a
volume, mass, area, or line by the method of integration, from the
equations of Art. 80 {Vx= I xdV, Ax= I xdA, etc.), it is possible
to select the element of volume, area, etc., in various ways and to
express the element in terms of either cartesian or polar coordi-
nates. The resulting integral may be a single, a double, or a triple
integral, depending on the way the element is selected. The
integral, of course, is a definite integral, the limits of integration
depending on the boundary curve or surface of the figure or body.
In any case the element of volume, mass, area, or Hne must be
taken so that,
1. All points of the element are the same distance from the line
or plane about which moments are taken; otherwise, the distance
from the line or plane to the element will be indefinite. Or, so that,
2. The centroid of the element is known, in which case the
moment of the element about the moment axis or plane is the
product of the element and the distance of its centroid from the
axis or plane.
The centroids of some of the common figures (Unes, areas, and
volumes) will be found in the following illustrative problems.
ILLUSTRATIVE PROBLEMS
Find, by the method of integration, the centroids of the following figures
with respect to the axes indicated.
172. Arc of a Circle. — The radius
which bisects the arc will be taken
as the a;-axis (Fig. 188). By symmetry
the centroid lies on this axis. Hence
y=0. If r denotes the radius of the
arc and 2 a the subtended angle, then,
in terms of polar coordinates, the ele-
ment of arc, dL, and its distance x from
the y-axis are dL = rdd and x=r cos 6.
Thus, the element of arc is selected in
accordance with the first of the above
rules and x may be found as follows: Fig. 188.
dL=rd0
164
FIRST MOMENTS AND CENTROIDS
Lx= { xdL
f
-X
r cos e • rdd
Therefore,
cos edd=2r^ sin a.
2r2 sin a _ 2r^ sin a _ r sin a
L 2ra a
If the arc is a semicircle, that is, if a =90° =^ radians, then, x= — . That
is, the distance of the centroid of a semicircular arc from the center of the
circle is slightly less than two-thirds of the radius of the circle.
173. Area of a Triangle. — In accordance with the first of the above
rules the elements of area wOl be taken as strips parallel to the base of the
triangle (Fig. 189) . Since each element is bisected by the median drawn from
the vertex opposite the base, the centroid of each element, and hence of the
entire area, lies on this median. If x denotes the width of the strip, the area
of thestrip isdA=a:c?2/. Thus,
-/
Ay= I xydy.
From similar triangles, the relation between x and y is.
Hence,
Therefore,
7 =r or x=T{h — y).
h—y h h ^
Ay = ^ (\h-y)ydy^\hhK
- Ihh^ 1,
The centroid of a triangular area, then, is at a
distance of one-third of the altitude from the
base.
174. Sector of a Circle. First Method. — The element of area will be
selected in accordance with the first of the above rules as indicated in Fig. 190.
Since the area is symmetrical with respect to the x-axis, the centroid lies on
this axis and hence ^ = 0. The value of x may then be found from the equa-
tion.
VOLUME OF RIGHT CIRCULAR CONE
Ax= I xdA
Therefore,
p cos 6-pdpdd = ir^ sin a.
fr^ sin a fr^ sin a 2 r sin a
165
If a = 90° = ^ radians, that is, if the sector is a semicircular area,
_ 4r
Fig. 190.
Fig. 191.
Second Method. — In accordance with the second of the above rules, the
element of area will be selected as a triangle, as indicated in Fig. 191. The
area of the triangle is hrHd and the distance of its centroid from the y-SLxis is
|r cos 6. Hence, the moment of the triangle with respect to the i/-axis is
Ir^ cos ddd and x is obtained from the equation,
Ax
COS odd
Therefore,
= fr'^ sin «.
_ tr** sin a
X-
2 r sin a
3
175. Volume of a Right Circular Cone. — The axis of the cone will be taken
as the a;-axis (Fig. 192). By symmetry, ^=0 and i = 0. In finding x, a thin
lamina parallel to the base of the cone will be selected as the element of
volume, the centroid of the lamina being at its center at the distance x from
the ?/2-plane. Hence, 5 may be found as follows:
Vx--
XttzHx.
166
FIRST MOMENTS AND CENTROIDS
From similar triangles, the relation between x and z is,
z r r
- = r or z = tX.
X h h
Hence,
Vx='^ \xHx = iTr%\
Therefore,
Fig. 192.
176. Volume of a Hemisphere. — The axis of symmetry will be taken as the
2-axis (Fig, 193) . By symmetry x=0 and y=0. A thin lamina parallel to the
base will be selected as the element of volume, the centroid of which is at the
distance z from the xy~p\a,ne. Hence, the 2-coordinate of the centroid of the
volume of the hemisphere may be found from the equation,
Vz
= I zirxHz=Tr I z{r^—z^)dz = l-i
Jo Jo
Therefore,
|7rr3 S*"-
Fig. 193.
177. Parabolic Segment. — Let the seg-
ment be bounded by the x-axis, the line
h^x
x = a, and the parabola y^= — as shown in Fig. 194. A strip parallel to
the 2/-axis will be selected as the element of area, the area of the strip being
expressed by ydx. The area of the segment, then, is.
A= \ dA= I ydx = —p \ Vxda
J Jo ^ajo
To find x,
Ax= I xdA
Fig. 194.
Jxydx = —p I x^^dx=iba\
0 vajo
Therefore,
A |a6 5
AREA OF QUADRANT OF AN ELLIPSE
167
To find y, the same elementary strip will be selected, but since each poin^
of the element is not the same distance from the a;-axis, its moment must be
expressed as the product of the area of the strip and its centroidal distance,
y
|, from the x-axis. Thus,
Therefore,
J "a 1 1,2 fa
hy"ydx = - — I xdx = \hH.
0 2 ajQ
^ A iah 8 •
178. Area of Quadrant of an Ellipse. — The semi-axes of the ellipse will
be denoted by a and h (Fig. 195) and hence the equation of the ellipse is
A strip parallel to the ?/-axis will be selected as the element of area. From
the equation of the ellipse y may be expressed in terms of x by the equation,
Hence,
A= I ydx = ^ \ ^a^-xHx = \irab.
Jo ajo
To find x;
lTrab'X= I xdA
= I x-ydx=- I xVa^—x^dx=la%.
Jo aJo
Therefore,
^■n-ab Sir'
Fig. 195.
To find y, the same strip will be used for the element of area, its centroid
being at the distance ^y from the x-axis. Thus,
Therefore,
J ''a 1^2 fa
\y-ydx= — I {a^—x^)dx = \ab^.
0 2avo
PROBLEMS
179. A pulley having a thin rim is 2 ft. in diameter. How far from the
center of the pulley is the centroid of each half of the rim?
180. Find the centroid of a semicircular area, the radius of which is
9 in.
168
FIRST MOMENTS AND CENTROIDS
181. Show that the centroid of the surface of a right circular cone is on
the axis of the cone at a distance of |/i from the apex, where h is the altitude
of the cone.
182. The radius of the base of a right circular cone is 5 in. and its height
is 2 ft. Locate the centroid of (a) the volume of the cone; (6) the curved sur-
face area of the cone.
183. Show that the distance of the cen-
troid of the surface of a hemisphere from
the base is |r where r is the radius of the
hemisphere.
Fig. 196.
184. Locate the centroid of the area
included between the 7/-axis, the line y = h,
and the parabola 7j^ = — as shown in Fig. 196.
Select the element of area as shown in the
figure. Ans. x=Ti)a, y = lb.
186. A paraboloid is generated by rotating the parabola y^ = px about the
X-axis. Locate the centroid of the volume included between the paraboloid
and the plane x = a.
Ans. x=|a.
83. Centroids of Composite Figures and Bodies. — As noted in
Art. 80, if the centroid of a line, area, volume, or mass is known,
the moment with respect to an axis or plane is most easily found by
multiplying the hne, area, volume, or mass by the distance of the
centroid from the axis or plane. Thus, if a given line, area, vol-
ume, or mass can be divided into parts, the centroids of which are
known, the moment of the whole line, area, etc., may be found
without integrating by obtaining the algebraic sum of the moments
of the parts into which the line, area, volume, or mass is divided,
the moment of each part being the product of that part and the dis-
tance of its centroid from the hne or plane. Thus, for example, in
the case of a composite area, if ai, ^2, ^3, etc., denote the parts
into which the area A is divided and x'o, Xo", Xo'", etc., denote the
a:-coordinates of the centroids of the respective parts, then.
or
or,
Similarly,
(01+^2+^3+ . . . )x = aiXo'-\-a2Xo''-\-a3Xo"' +.
i:a'X = X(axo)y
Ax='2{axo).
Ay = X{ayo).
CENTROIDS OF COMPOSITE FIGURES
169
From similar equations the centroid of a composite line, volume^
or mass may be found. Thus,
L^=S(Zxo), Vx='^(vxo), Mx=^{mxo),
Ly=7:{lyo), Vy = ^{vyo), My = i:{myQ),
ILLUSTRATIVE PROBLEMS
186. Locate the centroid of the T-section shown in Fig. 197.
Solution. — If axes be selected as indicated it is evident from symmetry
that x=0. By dividing; the given area into areas a^ and a^ and by taking
moments about the bottom edge of the area, y may be fomid as follows:
__12X1+12X5
^ "6X2+6X2
= 3 in.
T
nr,.,
N j ai f-^
7T
Vo
Fig. 197.
Fig. 198.
187. Locate the centroid of the volume of the cone and hemisphere shown
in Fig. 198, the values of r and h being 6 in. and 18 in. respectively.
Solution. — The axis of syinn.etry will be taken as the ?/-axis. From sym-
metry then x = 0. By taking the x-axis through the apex of the cone as shown
the equation Vy^'^ivyo) becomes,
(hrr'h+^r')y = ^7rr%Xlh-^^r^(h-\-lr).
That is,
Therefore,
-_f/t2+2rfe + fr2
^ h+2r
^fX(18)2+2X6Xl8 + fX(6)»
18+2X6
= 16.2 in.
170
FIRST MOMENTS AND CENTROIDS
PROBLEMS
188. Locate the centroid of the channel section shown in Fig. 199.
Arts. i=0.79 in.
189. The radii of the upper and lower bases of the frustum of a right cir-
cular cone are ri and r-i and the altitude is a. Find the distance of the centroid
above the lower base.
r<-3.^
3
Fig. 199.
Fig. 200.
Fig. 201.
190. Locate the centroid of the shaded area shown in Fig. 200.
191. In Fig. 201 is represented a homogeneous solid which consists of a
hemisphere and a right circular cylinder from which a cone is removed.
Locate the centroid of the solid with respect to the axes indicated.
Ans. i = 6.45 in.
192. Locate the centroid of the segment of a circle as shown in Fig. 202.
In the expression for x make a =-, and see if the result agrees with the result
2
found in Prob. 174 for a semickcle.
Fig. 202.
WiQ. 203,
THEOREMS OF PAPPUS AND GULDINUS
171
193. Find the distance from the larger base of the centroid of the area-of
h 261+62
the trapezoid shown in Fig. 203.
Ans. y
3 61+62*
194. A slender steel rod is bent in the form shown in Fig. 204. Locate
the centroid of the rod with respect to the axes shown.
1
10
— ^"
t^
5
.x^
p
'
^
Fig. 204.
Fig. 205.
195. Fig. 205 represents the cross-section of the end post of a bridge.
The area of each channel section is 4.78 sq. in. Find the distance from the
top of the cover plate to the centroid of the section.
84. Theorems of Pappus and Guldinus. — I. The area of a sur-
face of revolution generated by revolving a plane curve about any
non-intersecting axis in its plane is equal to the product of the
length of the curve and the length of the path described by the
centroid of the curve.
Proof.— Let the curve AB (Fig. 206) be revolved about OX,
The area of the surface generated is given by the equation,
A= i 2'Kij-dL = 2'K j ydL = 2Try'L,
Fig. 206.
Fig. 207.
172
FIRST MOMENTS AND CENTROIDS
where y is the distance of the centroid of the curve from OX and L
is the length of the curve.
II. The volume of the soUd generated by revolving any plane
area about any non-intersecting Une in its plane is the product of
the area and the length of the path described by the centroid of the
area.
Proof. — Let the plane area A (Fig. 207) be rotated about the
axis OX. Each elementary area dA will generate a circular ring
the volume of which is 2TrydA and hence the entire volume gen-
erated is given by the equation,
y=
j 27rydA = 27r \
ydA=2TryA,
where V is the distance of the centroid of the area from OX.
ILLUSTRATIVE PROBLEMS
196. Find the surface area and the volume of the soUd generated by
rotating a circle of radius r about a line in the plane of the circle at a distance a
from its center, a being greater than r.
Solution:
A=2iraX27rr = 47r2ar,
y = 27raX7rr2 = 27r2ar2.
L
>
i
\
■^ 3-^^—^
'1 i
Fig. 208.
197. A V-shaped groove is turned out of a
cylinder as indicated by Fig. 208. Find the
volume of the material removed.
Solution. — The distance of the centroid of
the triangle, which generates the volume, from
the axis of the cylinder is 2.5 in. and the area
of the triangle is 1.5 sq. in. Hence,
F = 27rX2.5Xl.5
= 23.55 cu. in.
PROBLEMS
Solve the following problems by the theorems of Pappus and Guldinus :
198. Find the surface area and the volume of a sphere.
199. Find the volume of a cone.
CENTER OF GRAVITY AND MASS-CENTER
173
Fig. 209.
200. Find the 7/-coordinate of the centroid of the trapezoid shown in^FigT
209. Determine the volume of the frustum
of a cone generated by revolving the trapezoid
about the x-axis.
201. Find the volume of the ellipsoid gene-
rated by revolving the right half of the ellipse,
-2-|-?^ = l, about the i/-axis.
85. Center of Gravity and Mass-
Center. Center of Gravity. — As noted
in Art. 80, if the distance of a certain
point (centroid) in an area, volume,
etc., from a given line or plane is known, the moment of
the area, volume, etc., with respect to the Une or plane may
be found by multiplying the whole area, volume, etc., by this
(centroidal) distance. Similarly, the moment of the resultant
earth-pull on a body (weight of the body) with respect to an axis
may be found by multiplying the weight of the body by the dis-
tance of a certain point in the body from the axis. This point is
called the center of gravity of the body. The center of gravity
of a body, then, is a point in the body (or its extension) such that
if the body were conceived to be concentrated at this point the
moment of the weight of the body with respect to any axis would
be the same as the algebraic sum of the moments of the weights
of the particles of the body with respect to the same axis.
The weights of the particles of a body constitute a system of
parallel forces having points of appHcation which are fixed in the
body and it can be shown ^ that the resultant of such a force
1 Center of a System of Parallel Forces. // a system of parallel forces
have fixed points of application in a rigid body the resultant of the system will pass
through a fixed point in the body regardless of the direction of the parallel force
system. This point is called the center
of the system of forces.
In order to show that the resultant
of such a force system does pass through
a definite point in the body for all direc-
tions of the parallel force system, it is
sufficient to prove the proposition for
two such forces. In Fig. 210 are shown
two parallel forces P and Q which act
at points A and B of a rigid body (not
shown). Let it be assumed first that
.P
/
X
/
A
R=P + Q
/
A
/
R=P+Q
l^
/q.
H H'
Fig.
210.
the forces act in the direction indicated
174 FIRST MOMENTS AND CENTROIDS
system passes through a definite point in the body regardless of
the way in which the body may be turned. Hence the center of
gravity of a body is a definite point in the body through which the
weight of the body (earth-pull on the body) acts regardless of the
position of the body.
If, then, W denotes the weight of a body and w or dW denotes
the weight of any particle of the body the coordinates of which are
X, y, and 2, the coordinates x, y, and z of the center of gravity of
the body may be found from the equations :
Fx = 2iyx= j xdW,
Wy=^n)y= j ydW,
/'
by the full lines. If the point in which the resultant of P and Q intersects the
line AB be denoted by H it can be shown (Art. 24) that AH = —~--XAB.
Let it be assumed next that the action lines of P and Q are rotated through an
angle 90° — 9 as indicated by the dotted lines, and that the action line of the
resultant of the two forces in this position intersects the line AB in the point
H\ Let BJ and H'l be drawn perpendicular to the action lines of P and Q
and let the point where the resultant of P and Q intersects BJ be denoted by K.
It can be shown as before that,
JK = IH' = ^^BJ = -^AB sin d.
Hence in the triangle AIH',
AH' = i^ = ^-AB.
sin d P+Q
Since AH = AH' it is evident that H' coincides with H, and hence the resultant
of P and Q passes through a definite point H on the line AB regardless of the
direction of the two parallel forces.
It is obvious that the proof may be extended to any number of forces, for,
the resultant of P and Q, just found, may be combined with a third force, S,
which has a fixed point of application C (say) in the body, and in the same
manner it can be shown that the resultant of these two forces will pass through
a fixed point on the line HC and so on.
So far as the relations of the forces to the body are concerned it is obviously
immaterial whether the body remains fixed and the forces be rotated relative
to the body as discussed above, or whether the forces remain fixed in direction
and the body be rotated. Hence the center of a system of parallel forces which
have fixed directions and which have fixed points of application in a rigid
body may also be defined as the point in the body through which the resultant
of the forces will pass regardless of the way in which the body is turned.
CENTER OF GRAVITY AND MASS-CENTER
175
Mass-center. — As defined in Art. 82, the centroid of the mass
of a body is the point in the body (or its extension) such that if
the mass were conceived to be concentrated at this point, its
moment with respect to any hne or plane would be equal to the
algebraic sum of the moments of the masses of the particles of the
body with respect to the hne or plane. In discussions (particularly
in Kinetics) in which the relation between mass and weight are
involved, the term mass-center (or center of mass) is frequently used
instead of centroid of mass.
It will be shown in Part III (Kinetics) that the mass of a body
/ l^\
is proportional to its weight {M= — ) . It can be shown, therefore,
that the center of mass of a body may, for all practical purposes,
be regarded as the same point as the center of gravity of the body.
Thus the a;-coordinate of the mass-center is
-f
xdM
M
fxgdW
Now, if g be regarded as constant (and for all practical purposes it
may be so regarded), the last term in the above equation becomes
fxdW
—^ — , and this is the expression for the x-coordinate of the
center of gravity of the body.
ILLUSTRATIVE PROBLEM
202. Gravel is piled on a floor so that the pressure on the beams which sup-
port the floor increases in the direction
of the beams as shown in Fig. 211.
The intensity of pressure varies from
zero at the left end of the beam to a
maximum ofp,» = 800 lb. per foot at
the right end. What is the weight of
the gravel and how far from the left
end of the beam does the resultant
weight act?
Solution. — Let the intensity of pres- FiG. 211,
sure at any distance x from the left end
of the beam be px lb. per foot. This pressure may be assumed to be constant
176
FIRST MOMENTS AND CENTROIDS
over a length dx and hence the weight of gravel over this length is dW = pxdx.
But Px^^yVm- Therefore the total pressure on the beam (weight of gravel) is,
W-
f
pxdx
= r ^Pmdx = ^Xl = 7200lh.
Jo l 2
Hence, the total pressure of the gravel (its weight) is the same in magnitude
as it would be if the gravel were spread uniformly to a depth equal to one-half
that of the ni;iximum depth. The total weight, however, would then act at
the center of the be:i,m. whereas, according to the above distribution it must
act at a distance x such that,
dWx,
Therefore,
Wx= Cc
x=l.
This result might have been obtained, without taking the above detailed
steps, from the fact that the gravel may be conceived to be concentrated in a
plane (of triangular shape), the position of the center of gravity of the gravel
then being the same as the centroid of the triangular area (Prob. 173).
PROBLEMS
203. Two homogeneous spheres A and B, connected by a rod, are mounted
on a vertical axis as shown in Fig.
P „ 212. The weights of the spheres are
_>j 3"l<^ 1 1 ~^ — ^^p~ 201b. and 601b., respectively, and the
^^y^ - , )D weight of the rod is 10 lb. How far
M>^ ^q'i ^ ^ 1G" V$L/ from the axis is the center of gravity
I ^ "^ of the three bodies? Ans. 7.94 in.
204. Four bodies A, B, C, and D
1<IG. 212. ^YQ carried by a rotating shaft as
shown in Fig. 213. The weights of
the bodies are 20, 15, 10, and 8 lb., respectively, and the distances of
their centers of gravity from the axis of the shaft are 12, 6, 5, and 10 in.,
respectively. Find the center of gravity of the four bodies when in the
positions shown.
DETERMINING CENTROIDS OF AREAS
177
205. The density at any point of a slender rod varies directly as the square
of the distance of the point from the end of the rod. Show that the center
of mass of the rod is fZ from the given end.
Q-
s^^^$^
^^ I
(Do
Q
^i^^;^
■^^^^
3%
Fig. 213.
206. A beam is loaded with brick piled so as to produce the distribution
of pressure as indicated in Fig. 214. Find the center of gravity of the brick.
\
\
1^ y w A
V '' ^'
m
< 5^ >
Fig. 214.
207. Three particles of equal mass are placed at the vertices of a triangle.
Show that the mass-center of the particles coincides with the centroid of
the area of the triangle.
86. Graphical Method of Determining Centroids of Areas. —
If the boundary of an area is
an irregular curve which can-
not be represented by an equa-
tion, the centroid cannot be
determined by the method of
integration. In such cases,
however, the centroid may be
determined by a graphical
method which will now be
described. Let B B'B'B (Fig.
215) be an area, the centroid Fig. 215.
of which is to be found.
Draw two parallel lines XX and X'X' any convenient distance
178 FIRST MOMENTS AND CENTROIDS
h apart. Draw a line BB parallel to XX and project the
points B,B on the line X'X', thereby obtaining the points
C,C. Connect the points C,C to any point P on XX. Let
the points where the lines CP intersect BB be denoted by D,D.
Repeat this process for other lines B'B', B"B", etc., and locate
similarly, points D', D'; D", D") etc. Connect the points
D, D; D', D'; D'\ D"; etc., by a smooth curve. Let the area
enclosed by this curve be denoted by A'. Then, if the given area
be denoted by A, the distance of the centroid of the area from the
line XX is given by the following equation,
Proof. — If DD and CC are denoted by x^ and x respectively,
then from the similar triangles PDD and PCC the following rela-
tion is obtained,
X x' , ,
y = — or xy = hx .
h y ^
The distance, y, of the centroid of the given area from XX,
then, may be found a follows:
jy-xdy hix'dy
= ^A'
The areas A and A' maybe determined by the use of a planimeter
or any other convenient method and the value of y is then deter-
mined from the above equation. In a similar manner, by taking
any two Knes YY and Y'Y' and proceeding as above, the distance,
X, of the centroid from YY may be determined.
87. Determination of Center of Gravity by Experiment. —
When a body is irregular in shape, the center of gravity cannot be
determined by the method of integration since the limits of the
integral cannot be determined. The center of gravity of such a
body, however, may be determined by the following experimental
methods.
Method of Suspension. — If a body be suspended by a cord the
center of gravity is on the (vertical) line coinciding with the axis
of the cord. This statement follows from the fact that the two
DETERMINING CENTER OF GRAVITY BY EXPERIMENT 179
forces which hold the body in equilibrium (the upward tension in
the cord and the downward earth-pull) must be equal, opposite,
and coUinear, and the earth-pull or weight of the body of course
acts through the center of gravity of the body. Hence if a body
be suspended from each of two points, the center of gravity will be
located in each of two lines in the body and hence is at the point
of intersection of the two lines.
Method of Balancing. — If a body be balanced on a knife-edge
the center of gravity of the body will be in a vertical plane through
the knife-edge. Hence if the body be balanced on a knife-edge in
three different positions, three such planes in the body will be
located, and the center of gravity of the body is the point in which
the three planes intersect. A slightly different method involving
the same principle, appHcable particularly to large bodies of known
weights, consists in resting one end of the body on a line of sup-
port (knife-edge) and the other end on platform scales. By
weighing the reaction (pressure) on the scales and by measuring
the distance between the knife-edge and the point of support on
the platform scales the action line of the weight of the body may be
found, since it lies in a plane which divides the distance between
the knife-edge and the point of support into segments inversely
proportional to the magnitudes of the reactions at the supports.
By repeating the process for two other positions of the body,
three such planes may be found as in the above method, and
hence the center of gravity may be located.
CHAPTER VI
SECOND MOMENT. MOMENT OF INERTIA
§ 1. Moments of Inertia of Areas
88. Moment of Inertia of an Area Defined. — In the analysis of
many engineering problems as, for example, in determining the
stresses in a beam or column, expressions of the form | x^dA are
frequently met, in which dA represents an element of an area A,
and X is the distance of the element from some axis in, or perpen-
dicular to, the plane of the area, the limits of integration being such
that each element of the area is included in the integration. An
expression of this form is called the second moment of the area or the
moment of inertia of the area with respect to the given axis. The
moment of inertia of an area with respect to an axis in, or per-
pendicular to, the plane of the area may, then, be defined as the
simi of the products obtained by multiplying each element of
the area by the square of its distance from the given axis.
The term moment of inertia is somewhat misleading, since
inertia is a property of physical bodies, only, and hence an area
does not possess inertia. For this reason the term, second moment
of an area, is to be preferred, particularly when contrasting the
expressions of the form here discussed with expressions which were
defined as first moments of areas in Chapter V. It may be noted
that each term x^dA in the summation which represents the
moment of inertia of an area can be written in the form x(oixiA),
and hence represents the moment of the moment of an element of
area, that is, the second moment of the element. The term
moment of inertia, however, is very widely used, due to the fact
that the expression is of the same form as an expression to be dis-
cussed later (Art. 100) which is defined as the moment of inertia
of a body and which does have a physical significance.
The moment of inertia of an area with respect to an axis will
be denoted by I for an axis in the plane of the area and by J
for an axis perpendicular to the plane of the area. The particular
180
POLAR MOMENT OF INERTIA
181
axis (or direction of the axis) about which the moment of inertia
is taken will be denoted by subscripts. Thus, the moments of
inertia of the area A (Fig. 216) with respect to the x- and y-sixes
are expressed as follows:
and
dA,
L= xMA,
Units and Sign. — Since
the moment of inertia of an
area is the sum of a number
of terms each of which is
the product of an area and
the square of a distance, the Fiq_ 216.
moment of inertia of an area
is expressed as a length to the fourth power. If, then, the inch
(or foot) be taken as the unit of length, the moment of inertia
wiU be expressed as inches (or feet) to the fourth power (written
in.^ or ft.^). Further, the sign of each of the products x^dA is
always positive since x^ is always positive, whether x is positive
or negative, and dA is essentially positive. Therefore the moment
of inertia, or second moment, of an area is always positive. In
this respect it differs from the first moment of an area, which may
be positive, negative, or zero, depending on the position of the
moment axis.
89. Polar Moment of Inertia. — The moment of inertia of an
area with respect to a line
perpendicular to the
plane of the area is called
the polar moment of iner-
tia of the area and, as
noted in Art. 88, will be
denoted by J. Thus the
polar moment of inertia,
with respect to the 2:-axis,
of an area in the xy-plane
(Fig. 217) may be ex-
FiG. 217. pressed as follows:
182 SECOND MOMENT. MOMENT OF INERTIA
Jz= ijrHA
-/'
= jx^dA+ jy^dA,
Therefore,
Hence the following proposition may be stated :
The polar moment of inertia of an area with respect to any axis
is equal to the sum of the moments of inertia of the area with respect
to any two rectangular axes in the plane of the area which intersect
on the given, polar axis.
90. Radius of Gyration. — Since the moment of inertia of an
area
I i x'^dA or I rHA, etc.,) is four dimensions of length, it
may be expressed as the product of the total area, A, and the
square of a distance, k. Thus,
Ix= \y^dA=Ak,^
or
r^dA=AkJ^.
-f
The distance k is called the radius of gyration of the area with
respect to the given axis, the subscript denoting the axis with
respect to which the moment of inertia is taken. The radius of
gyration of an area with respect to a hne, then, may be defined as
a distance such that, if the area were conceived to be concentrated
at this distance from the given hne, the moment of inertia would
be the same as the moment of inertia of the actual or distributed
area with respect to the same hne.
From the equation 7y= | x^dA=Akj/^, it will be noted that
ky^, the square of the radius of gyration with respect to the
?/-axis, is the mean of the squares of the distances, from the
2/-axis, of the equal elements of area into which the given area may
PARALLEL AXIS THEOREM FOR AREAS
183
be divided, and that it is not the square of the mean of these~3is^
tances. The mean distance (x) of the elements of area from the
y-SLxis is the centroidal distance as discussed in the preceding chap-
ter. Hence Ax^ does not represent the moment of inertia of an
area with respect to the y-Sixis.
91. Parallel Axis Theorem for Areas. — If the moment of
inertia of an area with respect to a centroidal axis in the plane of
the area is known, the
moment of inertia with
respect to any parallel
axis in the plane may
be determined, without
integrating, by means
of a proposition which
may be established as
follows: In Fig. 218 let
FF be any axis through
the centroid, C, of an
area and let Y^Y^ be
any axis parallel to FF
and at a distance d therefrom. Further, let the moment of inertia
of the area with respect to the axis FF be denoted by / and the
moment of inertia with respect to Y'Y' by /. By definition then,
/= j(x-\-d)^dA
= j x^dA -\-2d \xdA +d^ CdA .
I=l-\-Ad^ since \ xdA-=Ax = 0,
Therefore,
Hence the following proposition may be stated:
The moment of inertia of an area with respect to any axis in the
plane of the area is equal to the moment of inertia of the area with
respect to a parallel centroidal axis plus the product of the area and
the square of the distance between the two axes. This proposition is
called the parallel axis theorem.
A corresponding relation exists between the radii of gyration of
the area with respect to two parallel axes, one of which passes
184 SECOND MOMENT. MOMENT OF INERTIA
through the centroid of the area. For, by replacing I by Ak^ and
/ by ^A;^ the above equation becomes,
Ak^ = Al?-{-A(P.
Whence,
where k denotes the radius of gyration of the area with respect to
any axis in the plane of the area and k denotes the radius of gyra-
tion of the area with respect to a parallel centroidal axis.
Similarly, for polar moments of inertia and radii of gyration, it
can be showna that,
J = J+A(P,
and,
k^ = k^+d^,
where J and k denote the polar moment of inertia and radius of
gyration, respectively, of the area with respect to the centroidal
axis and J and k denote the polar moment of inertia and radius of
gyration, respectively, of the area with respect to an axis parallel
to the centroidal axis and at a distance d therefrom.
92. Moments of Inertia by Integration. — In determining the
moment of inertia of a plane area with respect to a Une, it is pos-
sible to select the element of area in various ways and to express
the area of the element in terms of either cartesian or polar coor-
dinates. Further, the integral may be either a single or double
integral, depending on the way in which the element of area is
selected; the Hmits of integration are determined, of course,
from the boundary curve of the area. In any case, however,
the elementary area must be taken so that :
(1) AU points in the element are equally distant from the axis
with respect to which the moment of inertia is to be found, other-
wise the distance x in the expression x^dA would be indefinite.
Or, so that,
(2) The moment of inertia of the element, with respect to the
axis about which the moment of inertia of the whole area is to be
found, is known, the moment of inertia of the area then being
found by smuming up the moments of inertia of the elements.
Or, so that,
(3) The centroid of the element is known and also the moment
of inertia of the element with respect to an axis which passes
MOMENTS OF INERTIA BY INTEGRATION
185
through the centroid of the element and is parallel to the given
axis; the moment of inertia of the element may then be expressed
by means of the parallel axis theorem.
The moments of inertia of some of the simple areas will now be
found in the following illustrative problems.
ILLUSTRATIVE PROBLEMS
208. Determine the moment of inertia of a rectangle, in terms of its base
b and altitude h, with respect to (a) a centroidal axis parallel to the base;
(b) an axis coinciding with the base.
Solution. — (a) Centroidal Axis. — The element of area will be selected in
accordance with rule (1) above, as indicated in Fig. 219. The moment of
inertia of the rectangular area with respect to the centroidal axis, then, is,
:=/.^.A=n
y^bdy
= .^J>hK
dy
I, , , ,,v,,r,T^
dA^bdy
T
Fig. 219.
-<|«M
^2z:
'/-' ■ ."Zr-^
dy
X
dA-bdy
Fig. 220.
(6) Axis Coinciding with the Base. First Method. — The element of area
will be selected as indicated in Fig. 220. The moment of inertia of the rect-
angle with respect to the base, then, is,
h
=frdA==£
y^bdy
:bh^
186
SECOND MOMENT. MOMENT OF INERTIA
Second Method. — Since the moment of inertia of the rectangle with respect
to a centroidal axis is rt^h^} the moment of inertia with respect to the base
may be found from the parallel axis theorem (Art. 91). Thus,
Ib = Ix+A
©
■■ IhhK
209. Determine the moment of inertia of a triangle, in terms of its base
h and altitude h, with respect to (a) an axis coinciding with its base; (6) a
centroidal axis parallel to the base.
Solution. — (a) Axis Coinciding with the Base. — The elementary area
will be selected as shown in Fig. 221 . The moment of inertia of the area of the
triangle with respect to the base, then, is.
/&= {yHA= I y^xdy.
But, from similar triangles,
X h-y
Hence,
Therefore,
h h
b
x=-^(h-y).
h=^j; Cy'ih-y)dy
-#*'•
Fig. 221.
Fig. 222
(6) Centroidal Axis Parallel to the Base. — ^The centroidal axis parallel
to the base (axis XX) is shown in Fig. 222 (see Prob. 173). Using the parallel
MOMENTS OF INERTIA BY INTEGRATION
187
axis thoerem, the moment of inertia of the triangular area with respect tothe
centroidal axis is,
=-hhh^~\hhX^h^
210. Determine the moment of inertia of the area of a circle, in terms of
its radius r, with respect to an axis coinciding with the diameter; (a) using
cartesian coordinates; (b) using polar coordinates.
Solution. — (a) Cartesian Coordinates. — ^The element of area will be
selected as shown in Fig. 223. The moment of inertia of the circular area
with respect to the diameter, then, is.
= CyHA=C'.
y^xdy
Fig. 223.
Fig. 224.
(b) Polar Coordinates. — The element of area will be selected as shown in
Fig. 224. Hence,
lx= I y'dA
I I (p sm 0)2 pdpdd
Jo Jo
188
SECOND MOMENT. MOMENT OF INERTIA
p3 sin2 d dpdd
sin2 d do
"T^sy 1 4
211. Determine the polar moment of inertia of the area of a circle of radius
r with respect to a centroidal axis : (a) by integration ; (6) by use of the theorem
of Art. 89.
Solution. — (a) By Integration. — By selecting the element of area as
indicated in Fig. 225, the polar moment of inertia of the circular area is,
=0
p^rrpdp
Fig. 225.
= 1 4
(6) By Use of Theorem of Art. 89.—
Snce Ix and ly are each equal to ^xr*
(Prob. 210), the polar moment of inertia of
the area of the circle is,
= 1.4
212. Find the moment of inertia, with respect to the rc-axis, of the area
bounded by the parabola y^=2x, and the line a: = 8 in.
Solution. First Method. — The element of area will be selected in accord-
ance with rule (1) of Art. 92 as indicated in Fig. 226. The moment of inertia
of the given area with respect to the x-axis, then, is.
= Cy^dA
^2 C y^{S-x)dy = 2 Ty^ U-t\dy
~\^ lojo
136.5 in.-*
MOMENTS OF INERTIA BY INTEGRATION
189
Second Method. — The elementary area will be selected in accordance with
rule (2) of Art. 92 as indicated in Fig. 227. Since each elementary area is a
Y
/v^3^
-^
■
y^ dx^
J
!>
-g 8—
II
<
>-
1
\
Fig. 226.
Fig. 227.
rectangle of width dx and height 2y, the moment of inertia of the element
with respect to the a;-axis is -^^dx {2yY=^yHx (see Prob. 208). Hence, the
moment of inertia of the given area is
/. = ^
. CyHx^'^- (^{2xf''dx
Jo 3 Jo
3 Jo
z^dx
.4V2
3
[h:
136.5 in.*
PROBLEMS
213. Determine the moment of inertia of the area of a circle, with respect
to an axis tangent to the circle, in terms of, r, the radius of the circle.
214. Determine the polar moment of inertia of the area of a rectangle of
base h and altitude d with respect to the centroidal axis.
Ans. J=-^bdib^+d^).
215. Find the moment of inertia and radius of gyration of a circular area,
16 in. in diameter, with respect to a diameter.
216. Determine the moments of inertia of the area of an ellipse, the prin-
cipal axes of which are 2a and 26, with respect to the principal axes.
Ans. /a = j7ra6^ Ib = hrba^.
190
SECOND MOMENT. MOMENT OF INERTIA
217. The base of a triangle is 8 in. and its altitude is 10 in. Find the
moment of inertia and radius of gyration of the area of the triangle with
respect to the base.
218. Find the polar moment of inertia and radius of gyration of the area
of a square, each side of which is 15 in., with respect to an axis through
one corner of the square.
219. Find the polar moment of inertia, with respect to a centroidal axis,
of the area of an isosceles triangle having a base b and altitude h.
Ans. J'=-iV6/i(i62+i/i2).
93. Moments of Inertia of Composite Areas. — When a compos-
ite area can be divided into a number of simple areas, such as tri-
angles, rectangles, and circles, for which the moments of inertia
are known, the moment of inertia of the entire area may be
obtained by taking the sum of the moments of inertia of the several
areas. Likewise, the moment of inertia of the part of an area
that remains after one or more simple areas are removed may be
found by subtracting, from the moment of inertia of the total area,
the sum of the moments of inertia of the several parts removed.
ILLUSTRATIVE PROBLEMS
220. Locate the horizontal centroidal axis, XX, of the T-section shown in
Fig. 228 and find the moment of inertia of the area with respect to this cen-
troidal axis.
Solution. First Method. — The dis-
tance, y, of the centroid of the area
from the axis XiXi may be found from
the equation,
■^
Ui
Xx-
:f
— X
Thus.
Ay = 'E(ayo).
_ 12X7+12X3
y-
12+12
K
= 5 in.
The moment of inertia with respect
to the XX axis is the sum of the
moments of inertia of the three parts
oi, 02, and as, with respect to that axis.
Thus,
Fig. 228.
7x=i^X6X(2)3+12X(2)2+iX2X(l)3 + ^X2X(5)3
=4+48 + . 67+83. 33
= 136 in.*
MOMENTS OF INERTIA OF COMPOSITE AREAS
191
Second Method. — The moment of inertia of the T-section may also be
determined as follows : First find the moment of inertia of the T-section with
respect to the axis XiZi by subtracting the moments of inertia of the parts
ai and at, from the moment of inertia of the rectangular area ABCD and then
find Ix for the T-section by use of the parallel axis theorem. Thus, the
moment of inertia, h, of the T-section with respect to the XiXi axis is,
and
/^ = iX6X(8)3-2X|X2X(6)3 = 736 m.\
Ix = I x-Ad^ ^736 -24:X(5y = lSQ in."
221. Find the moment of inertia of the channel section shown in Fig. 229,
with respect to the line XX. Find also the moment of inertia with respect
to the parallel centroidal axis.
X-
r
12^
Fig. 229.
Solution. — The area may be divided into triangles and rectangles as shown
in the figure. The values used in the solution may be put in tabular form as
shown below, where a denotes the area of any part, yo the distance of the cen-
troid of the part from the line XX, h the moment of inertia of the part with
respect to its own centroidal axis parallel to XX, and I'x the moment of inertia
of the part with respect to the axis XX.
Part.
a.
2/0.
ayo.
/o.
ayo^.
rx = Io+ayo^
ai
.745
1.61
1.20
.44
1.93
2.37
02
.745
1.61
1.20
.44
1.93
2.37
ai
.585
1.17
.68
.23
.80
1.03
Qa
.585
1.17
.68
.23
.80
1.03
as
3.360
.14
.47
.02
.07
.09
6.02 in. 2
4.23 in.3
6.89 in.*
Thus the moment of inertia Ix of the area with respect to the XX axis is,
/x = srj = 6.89in.'»
Further, the total area is A = Sa = 6.02 in. 2 and the moment of the area with
respect to the XX axis is S(a.vo)=4.23 in.^ Hence, the distance, y, of the
centroid of the area from the XX axis is,
y =
S(ayo)_4.23
A 6.02
=0.70 in.
192
SECOND MOMENT. MOMENT OF INERTIA
Therefore, the moment of inertia with respect to a line through the centroid
and parallel to XX is given by the equation,
Ix = Ix-Ad''
= 6.89-6.02X(.70)2
= 6.89-2.95
= 3.94 in."
PROBLEMS
222. A wooden column is built up of four 2-in. by 8-in. planks as shown
in Fig. 230. Find the moment of inertia of the cross-section with respect to
the centroidal axis ZX. Ans. /a; = 981 in.<
;%;-
l-< 8^^ ^
Fig. 230.
X-
\
.^n H ^
Y
Fig. 231.
223. Find the moment of inertia of the
angle section (Fig. 231) with respect to each of
the centroidal axes parallel to the two legs of the
angle.
224. In Fig. 232 is shown the cross-sot fion
of a standard 9-in. 21-lb. I-beam (fillets are
neglected). Find the moments of inertia of
the section with respect to the centroidal axes,
ZXand YY. Ans. /x = 84.9 in."; /j, = 5.16 in."
225. In Fig. 233 is shown the cross-section
of a standard 3 1 -in, by 5-in. standard Z-bar
(fillets are neglected). Find the moments of
inertia of the section with respect to the cen-
troidal axes XX and YY .
226. In Fig. 234 is shown a built-up
section made of a |-in. by 20-in. plate and
MOMENTS OF INERTIA OF COMPOSITE AREAS 193
four angles. Find the moment of inertia of the section with respert tn
the XX axis.
Ans. /x=1850 in.*
1 i
-U
^
5
i
1
•"—"^A
'
Vi
-e—
— \
■^
Fig. 233.
Fig. 234.
227. In Fig. 235 is represented a 16-in. circular plate in which there are
drilled five 2-in. holes and one 4-in. hole as shown. Find the moment of
inertia of the area of the holes with respect to the XX axis and also with
respect to the YY axis.
Ans. /a; = 252 in.*
X
_A._ . X
194
SECOND MOMENT. MOMENT OF INERTIA
228. Find the moment of inertia of a trapezoid (Fig. 236), in terms of its
bases and altitude, with respect to (a) an axis coinciding with its larger base;
(6) a centroidal axis parallel to the bases.
-br
-6o-
FiG. 236.
94. Product of Inertia Defined. — If the moments of inertia
of an area with respect to any two rectangular axes are known, the
moment of inertia with respect to any other axis through the point
of intersection of the two axes may frequently be obtained most
easily in terms of the moments of inertia of the area with respect
to the two rectangular axes and an expression of the form I xydA
in which dA is an element of the given area and x and y are the
coordinates of the element with respect to the two rectangular
axes. This expression is called the product of inertia of the area
with respect to the axes and is denoted by Pxy. Hence, the product
of inertia of an area with respect to any two rectangular axes
may be defined as the sum of the products obtained by multiplying
each element of area by the product of the two coordinates of the
element with respect to the two rectangular axes. That is,
=/
xydA,
The product of inertia of an area, Hke the moment of inertia of
an area, is of four dimensions in length and is therefore expressed
as inches (or feet, etc.) to the fourth power (in.^, ft.^, etc.).
Unhke moment of inertia, however, the product of inertia of an
area is not always positive, but may be negative or may be zero,
since either xory may be negative and hence their product may be
negative or equal to zero.
95. Axes of S3mimetry. — The product of inertia of an area with
respect to two rectangular axes is zero if either one of the axes
is an axis of symmetry. This follows from the fact that for each
AXES OF SYMMETRY
195
product xydA for an element on one side of the axis of sjonmetry
there is an equal product of opposite sign for the corresponding
element on the other side of the axis and hence, the expression
/
ocydA equals zero.
ILLUSTRATIVE PROBLEM
229. Find the product of inertia of the area of the triangle, shown in Fig.
237, with respect to the x- and y-a,xes.
Solution. — The relation be-
tween X and 7j is given by the
equation y = ^x. Hence,
xydA= I I xydxdy
= - I xHx = 12m.*'
dA=dXidy t
Fig. 237.
PROBLEMS
230. Find the product of inertia of the area of a rectangle, having a
and an altitude h, with respect to two adjacent sides.
231. Find the product of inertia of the quadrant of a circular area, shown in
Fig. 238, with respect to the x- and y-axes, in terms of its radius r.
Ans, Pxy = ir*.
Fig. 238.
Fig. 239.
232. Find the product of inertia of the rectangular area with respect to the
X- and ?/-axes as shown in Fig. 239.
196
SECOND MOMENT. MOMENT OF INERTIA
^,
>I<
1"
1
A
h -
1 •
1
' \
\
I ~
/
i
-X
233. Find the product of
inertia of the triangular area
with respect to the x- and y-axes
as shown in Fig. 240.
Ans. Pa
192 in.-*
Fig. 240.
234. Find the product of
inertia, with respect to the
coordinate axes, of the area
bounded by the parabola
y^ = ax, the line x = 'b, and the
a;-axis.
Ans. Pxv = \ah^.
96. Parallel Axis Theorem for Products of Inertia.— When
the product of inertia of an area is known for any pair of rectangular
axes passing through the centroid of the area, the product of
inertia of the area with respect to any parallel set of axes may be
determined without in-
tegrating. Thus, in Fig.
241, X'X' and Y'Y' are
axes which pass through
the centroid, C, of the
area; XX and YY are
parallel axes passing
through the point 0.
The coordinates of C
with respect to XX and
YY are denoted by x
and y. If the product
of inertia of the area with respect to XX and FF be denoted
by Pxv and the product of inertia with respect to X'X' and Y'Y'
be denoted by Pxy, then, by definition,
Y
Y'
X' _
|d!A\
y^
/
1
(
y
\
X-
C
I
\
i
Y'
/
/
y
^
Y
_^
y
Fig. 241.
.=J{x'+x)iy'+y)dA
= I x'y'dA+xy I dA-\-y I x'i
fy'c
dA-j-xi y'dA
Since each of the last two integrals is the first moment of the
area with respect to a centroidal axis, each integral is equal to
zero. The equation then becomes,
Pxv = Pxv+Axy.
PARALLEL AXIS THEOREM FOR PRODUCTS OF INERTIA 197
That is, the product of inertia of any area with respect to any pair
of rectangular axes in its plane is equal to the product of inertia
of the area with respect to a pair of parallel centroidal axes plus the
product of the area and the coordinates of the centroid of the area with
respect to the given pair of axes.
ILLUSTRATIVE PROBLEM
235. Find the product of inertia of the area shown in Fig. 242 with respect
to the X- and ^/-axes.
Solidion. — The area may be divided
into rectangles ai and Oa as shown. Using
the formula Pxy = Pxv-hAxy we have, for the
area ai,
Pxy = 0 + 12X1X3 = SQ in.4,
and for area 02,
p^ = 0+4X3Xl = 12in.<
Hence, for the entire area,
Pxj, = 36H-12 = 48in.<
PROBLEMS
I «o
0|
L
Fig. 242.
236. Find the product of inertia of the area shown in Fig. 243 with respect
to the X- and ^y-axes.
r^
->4< 4ii— >
-A
f
Fig. 243.
Fig. 244.
237. Locate the centroid of the angle section shown in Fig. 244 and deter-
mine the product of inertia with respect to centroidal axes parallel to the two
legs of the angle. Ans. P= — 18.3 in.-*
198
SECOND MOMENT. MOMENT OF INERTIA
97. Moments of Inertia with Respect to Inclined Axes. — The
moments of inertia of an area with respect to different Hnes (in
the plane of the area) which pass through a given point are in
general unequal. The determination of the moment of inertia of
an area by the method of
Y integration i s compara-
tively simple for certain
lines but rather difficult
for other hnes. Equations
(1) and (2) below make it
possible to determine the
moment of inertia with
respect to any line passing
through a given point in
the area in terms of the
moments of inertia and
product of inertia of the
area with respect to two rectangular axes passing through the
point. The equations may be derived as follows: The moment
of inertia of the area shown in Fig. 245 with respect to OX'
is expressed by the equation,
Ix'= I y'^dA = i (y cosd-x sin e)^dA
= cos2 d I y^d A +sin^ 6 | x'^dA — 2 sin 6 cos 6 I xydA
Fig. 245
Ix cos^d-\-Iy sin^ 6 — 2Pxy sin 6 cos
In a similar manner the following equation may be derived,
ly' = Ix sin^ S-\-Iy cos^ d-^2Pxv sin d cos 6.
(1)
(2)
Thus, from these equations, the moment of inertia of an area with
respect to an axis inchned at an angle d with one of a given pair
of rectangular axes may be found, without integrating, if the
moments of inertia and the product of inertia of the area with
respect to the given rectangular axes are known.
By adding equations (1) and (2) the following important equa-
tion is obtained,
ix' 1 'v' — 'z I •* y
(3)
PRINCIPAL AXES 199
That is, the sum of the moments of inertia of an area with respect"
to all pairs of rectangular axes, having a common point of intersec-
tion, is constant. It should be noted also that each side of equation
(3) is equal to the polar moment of inertia of the area with respect
to an axis passing through the point (Art. 89) .
Axes for Which the Product of Inertia is Zero. — It may be
shown that through any point in an area there is one set of rect-
angular axes for which the product of inertia is zero. Thus from
Fig. 245.
Pxv = (x'y'dA =J (^ cos d-\-y sin d){y cos d — x sin d)dA
= (cos2 d - sin2 d)JxydA + cos d sin df{y^ - x^)dA
=P^y COS 2e+\{I^-Iy) sin 20
2P
Hence Pxy = 0 when tan 26= j-^j '
ly J- X
98. Principal Axes. — In the analysis of many engineering
problems the moment of inertia of an area must be found with
respect to a certain axis called a principal axis. A ^principal axis of
inertia of an area, for a given point in the area, is an axis about
which the moment of inertia of the area is either greater or less
than for any other axis passing through the given point. It can
be proved that thiough any point in an area two rectangular axes
can be drawn for which the moments of inertia of the area are
greater and less, respectively, than for any other axes through the
point. There are then two principal axes of inertia of an area for
any point in the area. Further, it can be shown that axes for
which the product of inertia is zero are principal axes. And, since
the product of inertia of an area is zero for symmetrical axes, it
follows that axes of symmetry are principal axes. The above
statements may be demonstrated as follows:
The direction of the principal axes may be determined from
equation (1) of Art. 97 which may be written in the form,
Ir = Ix Fi rly 7. Pxv sm 26
H-cos2^ 1- cos 2(9
2 ^^' 2
lx\J-v I J^x -ly
COS 26 -P:cv sin 26.
XV
200
SECOND MOMENT. MOMENT OF INERTIA
The value of 6 which will make 7^' have a maximum or a minimum
value may be found by equating the first derivative of la^ with
respect to 6 to zero. Thus,
whence.
^ = sin 20(1 y- h) - 2P^ cos 26 = 0,
tan 20 =
2P^
L-h'
From this equation two values of 26 are obtained which differ by
180°, the corresponding values of 6 differing by 90°. For one value
of 6 the value of Ix^ will be a maximum and for the other, a min-
imum. If Pxv = 0 (which will always be the case if either the x- or
y-sixis is an axis of symmetry) the value of 6 is zero, and hence axes
of symmetry are principal axes.
ILLUSTRATIVE PROBLEM
238. Find the moments of inertia of the angle section, shown in Fig. 246,
with respect to principal axes passing through the centroid.
Solution. — The steps in the solution will be made as follows:
(1) The centroid of the area will be located, that, is x and y will be found.
(2) The moments of inertia and the product of inertia (ix, ly, and Pxy)
with respect to the centroidal x- and y-axes will then be found by the methods
discussed in Arts, 93 and 96.
(3) The directions of the
principal axes will then be found
by use of the equations of
Art. 98.
(4) The moment of inertia
with respect to each of the
principal axes, u and v, will then
be foimd by means of equations
(1) and (2) of Art. 97.
4XfX2+5|XfXT^
4Xt+5fXi
3.396 „_. .
= 3:6r=^^^^"-
4XfXA+5|XfX3A
^ 4Xf+5fXt
= ^=1.94 in.
Fig. 246.
PRINCIPAL AXES 201
/x=i^XtX(5|)3+5|XtX(U)' ^ ~ '
+TVX4X(t)'+4X|X(l|)2
= 5.57+3.30+0.02+4.59
= 13.48 in.*
/v=iVX5fX(f)'+5fXtX(f)2
+TVX|X4«+4XfX(l^)2
=0.02+1.19+2.00+1.69
=4.90 in.4
To determine Pxv, the value of F^fy will first be found and then the value of
Fxy may be found by means of the formula in Art. 96. Thus,
I xydxdy+ \ I xydxdy
3/i Jo Jo Jo
= 1.26+0.56
= 1.82 m.*
Using the formula of Art. 96 we have
= 1.82-3.61X(-0.94)X(-1.94)
= -4.76 in."
The directions of the principal axes are found from the formula of Art. 98,
Thus,
. ^o. _2X(-4.76)_
*^° 2^ = 4.90-13.48-^-^^'
/. 20 = 48° or 228°,
and
0 = 24° or 114°.
From the formula of Art. 97, the moment of inertia with respect to the axis
making an angle of 24° with CX (denoted by u) is,
7« = 13.48 cos2 24°+4.90 sin^ 24°-2(-4.76) sin 24° cos 24**
= 11.23+0.81+3.53
= 15.59 in.4
Using 0 = 114° and denoting the corresponding axis by v, we have,
/p = 13.48 cos2 114°+4.90 sin^ 114° -2( -4.76) sin 114° cos 114°
202 SECOND MOMENT. MOMENT OF INERTIA
= 2.23+4.08-3.53
= 2.78 in. 4
Hence, the principal moments of inertia are 15.59 in.^ and 2.78 in.'* The
corresponding radii of gjn-ation are 2.08 in. and 0.88 in.
PROBLEMS
239. In the Z-section shown in Fig. 247, 7a; = 25.32 in." and 7j, = 9.11 in.*
Find the principal moments of inertia. Ans. 31.2 in.*; 3.20 in.'*
240. Show that the moment of inertia of the area of a square is constant
for all axes in the plane of the area which pass through the center.
i±
-dVr^
s
/^ %"
Fig. 247.
241. Fig. 248 represents the cross-section of a standard 10-in. 25-lb.
I-beam. 7x = 122.1 in.", 7^ = 6.89 in.", and A =7.37 in. 2 Find the moment
of inertia and radius of gyration of the section with respect to a line making an
angle of 30° with the x-axis." Ans. 7 = 93.3 in.*
99. Moments of Inertia of Areas by Graphical and Approximate
Methods. — It is sometimes necessary to determine the moment
of inertia of an area having a bounding curve which cannot be
defined by a mathematical equation. The moment of inertia of
such an area may be determined by a graphical method or by an
approximate method.
MOMENTS OF INERTIA OF AREAS
203
Graphical Method. — Let the area within the outer curve in Fig.
249 be denoted by A and let the Une with respect to which the mo-
ment of inertia of A is to be determined be denoted by XX. Draw
any line, such as X'X', parallel to and at the distance h from XX.
Draw any chord, such as BB, and project the points B,B on X'X'
thereby obtaining the points C,C. Connect the points C,(7 to
any point, P, on XX by lines intersecting BB at points D,D.
Project the points D,D, on X'X' thereby obtaining the points
E,E. Connect the points E,E, to P by lines cutting the line
.k£-
I 1-^
c
E
-< 00^ 5-
1
E
c
B>
^
\
/
(
'
B ]
l(
A
\ ^' //
01
6
\
f
X
Fig. 249.
BB in points F,F, This construction may be repeated for a
number of chords parallel to BB, thus obtaining a number of
points similar to F,F. All such points may be connected by a
smooth curve, the area within which will be denoted by A'. The
moment of inertia of A with respect to the axis XX is given by the
product of the area of A' and h^ {A'h^). This fact may be proved
as follows: Denoting BB by x, DD by x', and FF by x" we have,
from the similar triangles PFF and PEE,
±1
y
h'
(1)
and, from the similar triangles PDD and PCCf
y h'
(2)
204 SECOND MOMENT. MOMENT OF INERTIA
From (1) and (2), the following equation is obtained,
The moment of inertia of the area A with respect to XX, then, is
given by the equation,
7= I yHA= I y^xdy = h^ I x"dy
= h^A\
The area A' may be obtained by means of a planimeter or by
approximate methods.
Approximate Method. — Although the method described above is
of importance, the amount of work involved in any numerical prob-
lem is considerable. An approximate method will now be described
which can easily be appUed to any area. For convenience, how-
ever, a simple area will be selected so that the approximate value
of the moment of inertia as determined by this method may be
compared with the exact value. Thus, let the moment of inertia
of the area of a rectangle, with respect to an axis coinciding with
its base, be found. The area may be divided into any convenient
number of equal narrow strips parallel to the base, as shown in
Fig. 250. (The narrower
^ the strips the more closely
will the result agree with
the exact result.) Let the
area be divided into ten
< 6- >i such strips each 0.2 in. in
Fig. 250. width. The moment of
inertia of the rectangle is
equal to the sum of the moments of inertia of the strips.
The moment of inertia of any particular strip with respect
to the base of the rectangle is
A-X6xaP+6XiX2/2,
where y is the distance of the centroid of the particular strip from
the base. The first term is small and may be omitted without
serious error. The moment of inertia of each strip then is approx-
MOMENT OF INERTIA OF MASS DEFINED 205
imately equal to the product of the area of the strip and the
square of the distance of its centroid from the base. Hence, the
moment of inertia of the rectangle is,
/ = f(.l2 + .32 + .52 + .72_^.92+l.l2_j_l,32_|.l_52_^172^192)
= fXl3.3
= 15.96 in 4
According to Art. 92 the exact value is,
/ = i5/i3 = ix6X23 = 16in.4
§ 2. Moments of Inertia of Bodies
100. Moment of Inertia of Mass Defined. — In the analysis of
the motion of a body, the body is frequently regarded as a system
of particles and expressions are frequently met in the analysis
which involve the mass of a particle and its distance from a
line or plane. The product of the mass of a particle and the first
power of its distance from a line or plane is called the first moment
of the particle as discussed in Art. 79 while the product of the
mass of the particle and the square of its distance from the line
or plane is called the second moment of the mass of the particle
or more frequently the moment of inertia of the mass of the particle
(or briefly the moment of inertia of the particle) with respect to
the line or plane. The moment of inertia of a system of particles
(mass-system or body) with respect to a fine or plane is the sum
of the moments of inertia of the particles with respect to the given
line or plane. Thus, if the masses of the particles of a system
are denoted by mi, m2, ms, . . . and the distances of the particles
from a given line are denoted by ri, r2, rs . . . , the moment of
inertia of the system may be expressed as follows:
I = miri^+m2r2^-\-m3r3^-{- . . .
= Smr^.
If the mass system constitutes a continuous body the summation
in the above equation may be replaced by a definite integral, and
the expression for the moment of inertia of the body then becomes,
=/
/= r^dM,
206 SECOND MOMENT. MOMENT OF INERTIA
where dM represents an element of mass of the body and r is the
distance of the element from the given line or plane. The limits
of the integral must, of course, be so chosen that each element of
mass of the body is included in the integration. Therefore, the
moment of inertia of a body with respect to a line or plane may be
defined as the sum of the products obtained by multiplying each
elementary mass of the body by the square of its distance from the
given line or plane. The moment of inertia of the mass of a body
(or more briefly the moment of inertia of a body) has a physical
significance, since common experience teaches that if a body
is free to rotate about an axis, the farther from the axis the mate-
rial is placed, that is, the greater the moment of inertia of
the body becomes, the greater is the moment of the forces required
to give the body a prescribed rotation in a specified time. Thus,
if a rod is free to rotate about a vertical axis and carries two spheres
which may be moved along the rod, experience shows that the
farther from the axis the spheres are placed the greater is the
torque required to produce a definite rotation in a given time.
Units. — ^No special one-term name has been given to the unit
of moment of inertia of a body, hence, the units of mass and the
unit of length used are specified. Thus, if the mass of a body is
expressed in pounds and the dimensions of the body are expressed
in feet, the moment of inertia of the body is expressed in pound-
foot^ units (written Ib.-ft.^).' In engineering problems, however,
the pound is generally used as the unit of force, in which case mass
/ W\
is equal to force divided by acceleration {M = — J, as will be dis-
cussed in Arts. 139 and 141, and the unit of mass is, therefore, a
derived unit being expressed in terms of the units of force (pound),
length (foot), and time (second). Thus, the dimensional equation
W L FT^
for mass is — = -^-^7^:3= -7^. And since the unit of moment of
9 T^ L
inertia involves the units of mass and of length squared,, then in
engineering problems the unit of moment of inertia is also a
prp2
derived unit, the dimensions of which are M L^ = -j—XL^ = FT^ L,
that is, Ib.-sec^-ft. The name geepound or slug is sometimes used for
the engineer's unit of mass, the first because the unit of mass is the
mass of a body that weighs g (32.2 approximately) pounds, and
the second because mass is a measure of the sluggishness (inertia)
RADIUS OF GYRATION 207
of the body. Thus the moment of inertia of a body is some-
times expressed in geepound-foot^ or slug-foot^ units (geepound-
ft.2 or slug-ft.^). It is important, however, to keep in mind that
the unit of moment of inertia is a derived unit, being expressed
in terms of the fundamental units of force, length, and time.
101. Radius of Gj^ation. — It is frequently convenient to
express the moment of inertia of a body in terms of factors, one of
which is the mass of the whole body. Since each term in the
expression for moment of inertia as defined above is one dimension
in mass and two dimensions in length, the moment of inertia of a
body may be expressed as the product of the mass, M, of the whole
body and the square of a length. This length is defined as the
radius of gyration of the body and will be denoted by k. Thus,
the moment of inertia, I, of a body with respect to a given line or
plane may be expressed by the product Mk^, and hence.
=/
rHM = Mk^ or k
~w
The radius of gyration of a body with respect to any axis, then, may
be regarded as the distance from the axis at which the mass may
be conceived to be concentrated and have the same moment of
inertia with respect to the axis as does the actual (or distributed)
mass.
Viewed differently, the radius of gyration of a body with respect
to an axis is a distance such that the square of this distance is the
mean of the squares of the distances from the axis of the (equal)
elements of mass into which the given body may be divided (not
the square of the mean of the distances).
102. Parallel Axis Theorem for Masses. — If the moment of
inertia of a body with respect to an axis passing through its cen-
troid (center of gravity or mass-center) is known, the moment of
inertia of the body with respect to any parallel axis may be found,
without integrating, by use of the following proposition.
The moment of inertia of a body with respect to any axis is equal
to the moment of inertia of the body with respect to a parallel axis
through the mass-center of the body plus the product of the mass of the
body and the square of the distance between the two axes.
This proposition may be stated in equational form as follows :
I-^l+Md'^,
208
SECOND MOMENT. MOMENT OF INERTIA
where 7 denotes the moment of inertia of the body with respect to
an axis through the mass-center and / denotes the moment of
inertia with respect to a parallel
axis which is at a distance d from
the axis through the mass-center.
Proof. — Let Fig. 251 represent
the cross-section of a body con-
taining the mass-center, G. Further,
let the moment of inertia of the
body with respect to an axis
through G and perpendicular to
this section be denoted by 7 and
let the moment of inertia with
respect to a parallel axis through
the point 0 be denoted by I. The expression for 7, then, is,
Fig. 251.
\(x-{-dy+y^]dM
= f{x^+y^)dM+d^ CdM+2d jxdM,
Therefore,
1=1+ M<P
since
/
xdM=Mx = 0.
This theorem is frequently called the parallel axis theorem for
masses. A similar relation may be found between the radii of
gyration with respect to the two axes. Thus, if the radii of gyra-
tion with respect to the two parallel axes be denoted by k and k, the
above equation may be written.
Hence,
Mk^ = MJc^-\-Md:^,
103. Moments of Inertia with Respect to Two Perpendicular
Planes. — The determination of the moment of inertia of a body
MOMENTS OF INERTIA OF SIMPLE SOLIDS
209
with respect to a line is frequently simplified by making use of the
following theorem : The sum of the moments of inertia of a body with
respect to two perpendicular planes is equal to the moment of inertia
of the body with respect to the line of intersection of the two planes.
Proof. — If the moment of inertia of the body (Fig. 252) with
respect to xy- and x^-planes be denoted by Ixy and 7xa, respectively,
the expressions for the moments of inertia are,
and
-f-
Fig. 252.
X By adding these two equations
the resulting equation is,
'hv+hz= f{z^+y^)dM
104. Moments of Inertia of Simple Solids by Integration. — In
determining the moment of inertia of a body with respect to an
axis by the method of integration, the mass of the body may be
divided into elements in various ways, and either cartesian or polar
coordinates may be used, leading to a single, double, or triple
integration, depending on the way the element is chosen. The
elements of mass should always be selected, however, so that,
(1) All points in the element are equally distant from the
axis (or plane) with respect to which the moment of inertia is to be
found, otherwise the distance from the axis to the element would
be indefinite. Or, so that,
(2) The moment of inertia of the element with respect to the
axis about which the moment of inertia of the body is to be found is
known; the moment of inertia of the body is then found by
summing up the moments of inertia of the elements. Or, so that,
(3) The mass-center of the element is known and the moment
of inertia of the element with respect to an axis through its mass-
center and parallel to the given axis is known, in which case, the
moment of inertia of the element may be expressed by use of the
parallel axis theorem (Art. 102).
210
SECOND MOMENT. MOMENT OF INERTIA
The moment of inertia of some of the simpler solids are found
in the following problems.
Note: The symbol 8 will be used in the following pages to denote the
density of a body.
ILLUSTRATIVE PROBLEMS
242. Determine the moment of inertia of a homogeneous right circular
cylinder with respect to its geometrical axis.
SoliUion. — In accordance with the first of the above rules, the element of
mass may be selected as indicated in Fig. 253. The volume of this element is
hpdpdd, and if the density be denoted by 6, the mass of the element is
bhpdpde. Hence, the expression for the moment of inertia becomes,
/= I pHM= ( ' \ "^ p^ bhpdpde
= hTr8hr*.
dA'pdddp
M
Z
rdk^dxdy
/
V:-
— r'
y
/
liil
1 '.
X
v/
/
/
/
C
/
/
%
/
/
N
a
Fig. 253.
Fig. 254.
Since the mass, M, of the cylinder is bi^r^h, the above equation may be
written,
_ 1
243. Determine the moment of inertia of a homogeneous rectangular
MOMENTS OF INERTIA OF SIMPLE SOLIDS 211
parallelepiped with respect to (a) a central axis parallel to an edge; (6)
a median line of one face.
Solution, (a) Central Axis Parallel to an Edge. — Let the lengths of
the edges be denoted by a, h, and c, and let the 2-axis, parallel to the edge c,
be the central axis about which the moment of inertia is to be found, as shown
in Fig. 254. The parallelopiped may be divided into elementary volumes as
indicated, the mass of each- element being Scdxdy. The moment of inertia
with respect to the 2-axis, then, is,
rHM-= I I 8c(x^-\-y^)dxdy
J_a J J)_
2 2
2
, (a^h , ahA
In a similar way it can be shown that the moments of inertia with respect to
central axes parallel to the edges a and h are
and
/:r = ^M(62+c2)
/„ = — M(a2+c2).
Thus, the moment of inertia of a homogeneous rectangular prism with
respect to a central axis parallel to any one of the three edges is expressed in
terms of the mass and the lengths of the other two edges.
(6) A Median Line of One Face. — In order to determine the moment of
inertia with respect to a median line of one face the formula of Art. 102 may
be used. Thus, if the line MN (Fig. 254) be taken as the axis, the moment of
inertia with respect to MN is
lMN = Iz^^Mi^*
=iVM(a2+62)+iMa2 = iM(4a2+62).
244. Determine the moment of inertia of a homogeneous sphere with
respect to a diameter.
212
SECOND MOMENT. MOMENT OF INERTIA
Solution. — The cross-section of the sphere in the xy-plane is shown in
Fig. '255. In accordance with the second rule above, the element of volume
may be taken as a thin circular lamina included between two planes parallel
to the xz-plane, as shown in cross-section. This element, then, may be
regarded as a circular cylinder of radius x and altitude dy. The mass of the
elemental cylinder is birxHy and its moment of inertia with respect to the ?/-axis
is \birxHy (Prob. 242). Hence, the moment of inertia of the entire sphere
with respect to the y-axis is,
x*dy
= i^57rr6
= l(|57rr3)r2
y^)Hy
Fig. 255.
5
245. Determine the moment of
inertia of a homogeneous right
circular cone about its geometrical
axis.
Solution. — A cross-section of the cone in the xy-p\a.ne is shown in Fig. 256,
the y-a,xis being taken as the geometric axis. In accordance with the second
rule above, the element of volume is taken as a circular lamina included
between two planes parallel to the base, the height of the lamina being d]i.
The mass of this elemental cylinder is SirxHy and its moment of inertia with
respect to the y-axis is ^Sirx^dy. The moment of inertia of the cone with
respect to the y-&xia, then, is,
= ^57r j
x*dy.
The relation between x and y may be
found from similar triangles. Thus,
h^_ h—y
r X
That is,
x=j^{h-y).
Hence,
ly^W^.C {h-yYdy
MOMENTS OF INERTIA OF SIMPLE SOLIDS
213
Second Method. — If the element of volume be taken in accordance with the
first rule above, as indicated in Fig. 257, the element of mass then is bzpdpdd
and hence the moment of inertia of the cone with respect to the z-axis is,
h
iP
bzpHpdd.
The relation between z and p may be
found from similar triangles. Thus,
g Ji
r — p r
Hence,
2 = -(^-p)-
XT'
_8h
r
■ToSTrhr*
^{r-p)dpdd
= lMrK
^dk^pdddp
Fig. 257.
246. Show that the moment of inertia of a homogeneous right prism,
having a cross-section of any shape, with respect to a -plane coinciding with
one of the bases of the prism, is / = \Ml^ in which M is the mass of the rod and
I is the length of the rod. Further, show that for a slender rod the above
expression represents, with slight error, the moment of inertia of the rod with
respect to an axis passing through one end of the rod and perpendicular to
the rod.
Fig. 258.
Solution. ^In accordance with the first of the above rules, the element of
mass may be selected as a thin lamina parallel to the end plane as shown in
214
SECOND MOMENT. MOMENT OF INERTIA
Fig. 258. The volume of the element is Adx and its mass is SAdx, where A is
the cross-section of the prism. Hence,
'fy
I = 8A I xHx
= ^-^ = \{bAl)P = \Ml\
If the moment of inertia is taken with respect to an end axis {Y Y), then, x
(Fig. 258) is not the distance of all points in the lamina from the axis; the
greater part of the lamina is farther from the axis than the distance x. There-
fore the product dM •x^ = {bAdx)x^ is less than the real moment of inertia of
the lamina, and hence \Ml^ is less than the true moment of inertia of the rod
with respect to the end axis YY . But, the products dMx^ for the laminas
which are greatly in error (for the laminas near the axis YY) have little influence
in determining the value of the summation, | dMx^, in comparison with the
laminas far removed from the axis and for which the products dMx^ are only
slightly in error. Hence, the expression ^Ml- is a close approximation to
the moment of inertia of a slender rod, as for example, a flywheel spoke or
connecting rod, etc., with respect to an axis through an end. Thus, the error
in the expression ^Ml^ is only 1 per cent for a circular cylinder when the
length of the cylinder is only 4.3 times its diameter.
247. Show that the moment of inertia of a homogeneous thin circular
lamina with respect to an axis through the mass-center, parallel to the bases
of the lamina, is approximately IMr^, in which M is the mass of the lamina
and r is the radius.
Solution. — ^A top view and an end view of the lamina are shown in Fig. 259.
The element of volume will be taken as a prism of altitude t and cross-section
pdpdd, as indicated in the figure. The mass of the element is btpdpdd . Now
if the thickness t is relatively small, all points in any elementary
prism are approximately at the same distance {y = p sin d) from the
X-axis except for those prisms which are near the axis, and these prisms
contribute little to the moment
of inertia of the lamina with
respect to the a:-axis. Thus, the
moment of inertia of the lamina
with respect to an axis through its
mass-center parallel to the bases of
the lamina is approximately,
-XX
Up^ sin2 edpdd
Fig. 259.
= \UTrr* = \{bt-Kr'^)r^
4
MOMENTS OF INERTIA OF SIMPLE SOLIDS
215
It should be noted that the smaller the value of t becomes, the closer is the
approximation. The above expression is also a close approximation to the
moment of inertia of the lamina with respect to a diameter of either base of
the lamina.
248. Determine the moment of inertia of a homogeneous right circular
cylinder with respect to a diameter of one of the bases.
Solution. First Method. — In accordance with the third of the above rules,
the element of volume may be taken as a thin circular lamina parallel to the
base as indicated in Fig. 260. The mass of the element is b-wrHz and the
moment of inertia of the element with respect to a centroidal axis parallel to the
X-axis, as found in the preceding problem, is \b-Kr'^dz. The moment of inertia
of the element with respect to the a:-axis, then, as found by the parallel axis
theorem (Art. 102), is
\birrHz-\-bTrr^zHz.
Hence, the moment of inertia of the entire cylinder with respect to the x-axis is,
/= j \bTrrHz-\- j birr^zHz
= i57rr*/i+i57rr2A3
= ii257rr2;i(3r2+4/i2)
= ^M(3r2+4/i2).
Fig. 260.
Fig. 261.
Second Method. — The moment of inertia with respect to the a;-axis may
also be found by adding the moments of inertia with respect to the xy- and
xz-planes (Art. 103). Thus, in Fig. 261,
216
SECOND MOMENT. MOMENT OF INERTIA
From Problem 246 the moment of inertia with respect to the end (xy) plane is
To find the moment of inertia with respect to the xz-plane an element of
may be selected as indicated in Fig. 261. Thus,
;= ( yHM= I \%bh2xdy)
= 2bh \ yWr'^-yHy
Therefore,
249. Determine the moment of inertia of a homogeneous right circular
cone with respect to a diameter of the base.
Solution. — In accordance with the third of the above rules the element of
volume may be taken as a thin cylindrical lamina parallel to the base, as indi-
cated in Fig. 262. The mass of this element is birxHz and its moment of
inertia with respect to its centroidal axis parallel to the x-axis is ^hnrx'^dz (see
Problem 247). The moment of inertia with respect to the x-axis, according
to the parallel axis theorem (Art. 102), is \bTrxHz-^bTrx'^zHz. Hence, the
moment of inertia of the entire cone with respect to the a:-axis is,
Ix= y\bitx'dz+ {
From similar triangles,
r
x=~(h — z).
n
8irx^z^dz.
Hence,
Ix
iSTT- I {h-zYdz^h-K~X ^Kh-zydz
Fig. 262.
= i57rr2A(^r2+-iVA2)
=^M{Sr^+2h^).
250. Determine the moment of inertia of a sphere with respect to a central
axis if the density at any point varies directly as the distance of the point from
a central plane perpendicular to the axis.
MOMENTS OF INERTIA OF SIMPLE SOLIDS
217
Solution. — The moment axis will be taken as the y-axis of coordinates
(Fig. 263). The element of volume will be taken as a thin lamina parallel
to the a:z-plane. The density at all points in this lamina is proportional to y
and hence may be>denoted by ky. Hence, the element of mass is kyirx^dy
and the mass of the sphere is
M
f + r r+r
= kir \ yxHy = k'K I (r^ — y^)ydy = ^Trkr^
The moment of inertia of the cylindrical lamina is ^kyirxHy (Prob. 242)
and the moment of inertia of the entire sphere is the suln of the moments of
inertia of the elementary cylinders. Thus,
yxHy
y{r^-y^Ydy
Ik-wr^
-J(^/e7rr4)r2
Mr^
Fig. 263.
251. Find the moment of inertia, with respect to a central axis, of a cast-
iron sphere 10 in. in diameter. Assume the weight of cast iron to be 450 lb.
per cubic foot.
Solution. — The mass of the sphere is.
^ 450 . ^- lb .-sec. 2 ,
X?^?r-?,=4.27 — jz or slugs.
ft.
^=3'^ll2J '^32.2
5
= .296 lb.-sec.2-ft. (or slug-ft.2).
PROBLEMS
252. Determine the moment of inertia of a homogeneous right circular
cone about an axis through the center of gravity perpendicular to the geo-
metrical axis. Ans. T6M(r^-{-lh^).
218 SECOND MOMENT. MOMENT OF INERTIA
253. A homogeneous cylindrical rod is 2 ft. long and the radius is 2 in.
If the moment of inertia is found with respect to a line through one end of the
rod perpendicular to its axis by using the approximate formula of Problem
246, what is the error in the result, in per cent?
254. Determine the moment of inertia of a steel cylinder 6 in. in diameter
and 12 in. high, about its geometric axis. Assume the weight of steel to be
490 lb. per cubic foot.
255. Show that the moment of inertia of a homogeneous slender rod with
respect to an axis through the mid-point of the rod and perpendicular to the
rod is approximately -^Ml"^.
256. Determine the moment of inertia of a homogeneous ellipsoid, the
principal axes of which are 2a, 26, and 2c, about the axis 2a.
Ans. / = iM(62+c2).
257. Determine the moment of inertia of a homogeneous elliptic cylinder,
in which the principal axes of the cross-section are 2a and 26, with respect to
(1) the geometrical axis, and (2) an axis through the center of gravity coin-
cident with the axis 2a of the cross-section.
Ans. (1) 7 = iM(a2+62); (2)7 = -jLM(/i2+362).
105. Moments of Inertia of Composite Bodies. — If a body can
be divided into several finite parts, the moment of inertia of each of
which is known, the moment of inertia of the given body may be
obtained by adding the moments of inertia of the several parts.
In like manner if parts of a body are removed, the moment of
inertia of the remainder may be obtained by subtracting from the
moment of inertia of the original body the sum of the moments
of inertia of the parts removed.
ILLUSTRATIVE PROBLEMS
258. Determine the moment of inertia of a homogeneous, hollow, circular
cylinder with respect to its geometric axis, in terms of its mass M and its inner
and outer radii, n and r2.
Solution. — Let 1 2 and M^ denote the moment of inertia and the mass of a
solid cylinder of radius r2 and let Ii and Mi have similar meanings for the
cylinder of radius ri which is removed. Then,
I = h-h = hM2r2^-\Mirr'
= I Sxra^/i - 1 SttTi^/i = i 57r/i(r2* - ri^)
= i57rA(r22-ri2)(r22-i-ri2) = ^Mir^+r^'').
£1
MOMENTS OF INERTIA OF COMPOSITE BODIES 219
259. Determine the moment of inertia of the cast-iron flywheel shown in
Fig. 264 with respect to the axis of rotation. Assume the weight of cast iron
to be 450 lb. per cubic foot.
Fig. 264.
Solviion. — The rim and hub are hollow cylinders and the spokes may be
regarded as slender rods.
The weight of the rim =^^[(11) '" (x|) H Xj|x450 = 903 lb.
The weight of the hub ='rr (^) '- (^) H X^X450 = 172 lb.
■I r o 1 IT
The weight of each spoke =7r X-nr Xtt^ Xt^ X450 =41 . 7 lb.
LJi i.^ i.^
For then™. '-imirMWV''^'^'^-'''
For the hub, 7 = >x^[(|) +g)] =0.54 slug-ft.^
FortheBpokes, /=6X [j^xf-^X (g V|-^X i^^) "J = 11.1 slug-ft.'
Hence the moment of inertia of the flywheel is,
7 = 102.3+0.54 + 11.1 = 113.9 slug-ft.2
PROBLEMS
260. Determine the moment of inertia of the frustum of a homogeneous
right circular cone with respect to the geometrical axis, the radii of the bases
being r2 and ri. . ^ 3 ,^^2^-ri^
220
SECOND MOMENT. MOMENT OF INERTIA
261. The head of the mallet shown in Fig. 265 is a rectangular parallel-
epiped and the handle is a right circular cylinder. If the weight of the mate-
rial is J lb. per cubic inch, find the moment of inertia of the mallet with
respect to the line YY. Ans. / = 3.83 slug-ft.^
-20-
1
?3Z[)
Fig. 265.
Fig. 266.
262. The wheel shown in Fig. 266 is made of cast iron and has a solid web.
Determine the moment of inertia of the wheel with respect to the axis of
rotation. Assume the weight of cast iron to be 450 lb. per cu. ft.
263. From a round steel disk which is 20 in. in diameter and 4 in. thick, are
bored four holes, each 4 in. in diameter. The axes of the holes are parallel
to the geometric axis of the disk and 5 in. therefrom. Find the moment of
inertia of the remainder of the disk with respect to its geometric axis, assuming
the weight of steel to be 490 lb. per cu. ft.
A71S. / = 3.51 slug-ft.2
264. Two spheres are connected by a horizontal rod and are free to rotate
about a vertical axis midway between the spheres. The diameter of each
sphere is 9 in., the distance between their centers is 2 ft., and the diameter of
the rod is 2 in. Find the moment of inertia of the rod and spheres with respect
to the axis of rotation. Assume that the rod and spheres are made of cast
iron which weighs 450 lb. per cu. ft.
106. Moments of Inertia of Bodies by Experimental Methods.
— If a body is irregular in shape, the moment of inertia cannot be
found by methods of integration, since it is impossible to deter-
mine the Hmits of the integral. The moments of inertia of such
bodies may be determined experimentally, however, by methods
which make use of the laws of motion of a pendulum. These
methods of determining the moment of inertia are discussed
under the subject of Pendulums (Chapter IX).
PART II
KINEMATICS
107. Introduction. — Kinematics treats of the motion of bodies
without considering the manner in which the motion is influenced,
either by the forces acting on the bodies or by the character of
the bodies themselves. That is, the bodies are treated as geo-
metric soUds and not physical bodies. When the geometric solids
are endowed with physical properties, we are led to a study of
force, energy, momentum, etc., that is, to a study of Kinetics
(Part III).
Kinematics deals with the relation between distance, time,
velocity, and acceleration. In order to build up the fundamental
conceptions which are involved in the study of the motion of
bodies, the kinematics of a particle (material point) will be
treated first. A particle is a body, or part of a body, the dimen-
sions of which are negligible compared with its range of motion.
Bodies are made up of particles, and the study of the motion of
bodies is largely a study of the motion of their particles.
In treating of the motion of bodies (Chapter VIII), only rigid
bodies will be considered, and the motion of the rigid bodies will
be restricted to translation, rotation, and plane motion.
221
CHAPTER VII
MOTION OF A PARTICLE
108. Tjrpes of Motion. — The motion of a particle along a
straight line path is called rectilinear motion. The motion of a
particle along a curved path is called curvilinear motion. If the
moving particle describes equal distances in equal periods of time,
however small, the motion is said to be uniform. If unequal dis-
tances are described by the moving point in equal periods of time,
the motion is said to be non-uniform or variable.
Thus, if the crank shaft of a steam engine revolves at a constant
-X
Fig. 267.
number of revolutions per minute, the crosshead of the engine
has a non-uniform rectiUnear motion, the crank pin has a uniform
curvilinear motion, and any intermediate point on the connecting
rod has a non-uniform, curvilinear motion.
109. Linear Displacement. — The displacement of a moving
point is its change of position. The position of a moving point, at
222
LINEAR DISPLACEMENT 223
any instant, may be specified in a number of ways, as, for example,
by stating the rectangular coordinates or the polar coordinates of
the point. Thus, in Fig. 267 the position, at any instant, of the
point M as it travels along the curve from B to C may be specified
by the coordinates {x, y) or by (p, 6), p being the radius vector and
6 its direction angle. The displacement Ac of the point as it moves
from the position (xi, yi) or (pi, ^i) to the position (x2, 2/2) or
(p2, B2) is the straight fine jBC, that is, the vector drawn from
B to C. This displacement may be expressed as the vector sum^ of
its.T- and ^/-components. Thus,
Ac = Ao; 4> Ay,
in which, !
Ax = X2— a:i is the displacement in the a;-direction, and
A?/ = ?/2— 2/1 is the displacement in the ^/-direction.
The magnitude of Ac may be expressed algebraically by the
equation,
Ac=\/(Aa:)2+(Ai/)2,
and its direction with the x-axis may be expressed by the equation,
tan <t> = ~.
Ax
The displacement Ac may be expressed also as the vector dif-
ference of the radius vectors to the two positions of the moving
point. Thus,
Ac = p2— >pi,
that is, Ac is the directed distance which must be added to pi to
give p2. Or, in other words, p2 is the vector sum of pi and Ac.
1 The symbol +^ will be used in this and succeeding chapters, when dealing
with vector quantities, to denote the fact that the addition of the quantities
is a geometric or vector addition. Thus, if 72 is the resultant of the forces
P and Q, the triangle law may be expressed by the equation R = P-HQ. And,
if u and w are components of the velocity v, this fact may be expressed bj'- the
equation v = u-^ w. Similarly, the symbol -^ will be used to denote vector
subtraction. Thus, V2 -^ Vi denotes the geometric or vector difference of two
velocities and not their scalar difference, that is, the difference of their magni-
tudes, only.
224 MOTION OF A PARTICLE
Expressed algebraically, by the law of cosines, the magnitude of
Ac may be written,
Ac= Vpi^+pa^ — 2pip2 cos A^.
The unit of displacement is any convenient unit of length,
such as the inch, foot, mile, etc. It will be noted, however, that
displacement is a directed distance or length, that is, a vector
quantity. Displacements, therefore, may be combined and
resolved according to the parallelogram (or triangle) law Hke forces
and other vector quantities. It is important to note that by one
of the above equations the displacement is expressed as the vector
sum of two directed distances, whereas by the other equation it is
expressed as the vector difference of two directed distances. Fur-
ther, one vector equation is sufficient to express both the magnitude
and the direction of the displacement, whereas two algebraic equa-
tions are required for the same purpose.
If the displacement of the particle is decreased indefinitely,
the point C (Fig. 267) will approach the point B and, in the limit,
the chord Ac becomes coincident with the tangent to the path at B.
Therefore, the direction of motion of the particle at any point on
its path is tangent to the path at that point.
110. Angular Displacement. — The angular displacement of a
moving point is the change in the angle which the radius vector
to the point makes with some fixed reference line. Thus, in
Fig. 267, the angular displacement, A^, corresponding to the linear
displacement, Ac, is,
Ad= 61 — 62.
The unit of angular displacement may be any convenient
angular measure, such as the degree, revolution, radian, etc.
It is important to note that the angular displacement of a point
depends upon the reference point or pole about which the radius
vector is assumed to revolve. If the point moves on a circular
arc, the radius vector is usually taken as the radius of the circle
and the pole is the center of the circle.
LINEAR AND ANGULAR DISPLACEMENTS
225
PROBLEMS
265. An automobile starts from the position A, as specified with reference
to the axes shown in Fig. 268, and travels to position B along the road indicated
by the curved path. Find (a) the magnitude and direction of the linear dis-
placement of the automobile; (6) the angular displacement.
Fig. 268.
Fig. 269.
266. A point moves on a circular path of radius r from A to B (Fig. 269),
the angular displacement, with reference to the center, 0, of the circle, being
denoted by Ad and, with reference to the pole Oi, by A(f>. Show that the mag-
nitude of Ac, the linear displacement of the point, is expressed by the equation,
Ac = 2r sm — .
U Ad=2 radians (45°), what is angular displacement, A0, with respect to Oi?
111. Relation between Linear
and Angular Displacements. — If
a point moves along a circular
path, the linear displacement, dc
(Fig. 270), corresponding to an
indefinitely small angular dis-
placement do may be considered
to be coincident with the arc ds,
which is subtended by the angle
do. Since the arc of a circle is
the product of the radius and
the central angle, when the angle
is measured in radians, the rela-
ds^Ua
Fig. 270.
ds=de
f
pdd
//
y
A
226 MOTION OF A PARTICLE
tion between the linear and angular displacements may be ex-
pressed by the equation,
ds = rdd.
For a large angular displacement A^, the corresponding linear
displacement Ac is not equal
to rM. The distance As
along the arc, however, is
expressed by rM.
If the moving point does
not travel on a circular
path, that is, if the path
does not have a constant
radius of curvature, the
equation ds=rdd may be
used, provided that r is
the radius of curvature
Fig. 271. of the path at the given
position of the point and
that do is measured with respect to the center of curvature
as the pole.
If, however, the pole is not chosen as the center of curvature
of the path (Fig. 271), the displacement may then be expressed in
terms of its two components as follows :
ds = pdd -{^dp,
in which p is the radius vector to the point and not the radius of
curvature of the path at that point.
112. Linear Velocity and Speed. — The linear velocity of a
moving particle is the time rate at which the particle is changing
position, or, more briefly, the time rate of linear displacement.
The direction of the velocity of the moving particle at a given
point on its path is tangent to the path at that point (Art. 109).
Hence velocity, Hke displacement, possesses both magnitude and
direction. The magnitude of the velocity of a point is called the
speed of the point. Speed, therefore, is a scalar quantity. It may
be defined as the time rate of describing distance (not the time rate
of displacement). Although the terms velocity and speed are fre-
quently used interchangeably, it is important to associate wdth the
word velocity the two properties which it possesses, for, a change in
LINEAR VELOCITY AND SPEED 227
the direction of a velocity is fully as important in the laws of
motion of physical bodies as is a change in the speed.
If a point has a uniform motion along any path, the speed of
the point is the ratio of any distance, As, described by the point,
to the corresponding interval of time, Ai. Thus,
As
It will be noted that if a particle has uniform motion, whether
rectilinear or curvilinear, the speed of the particle is constant.
The velocity, however, is constant only in the case of uniform
rectilinear motion, since in any curvilinear motion the velocity of
the particle continually changes direction.
If the motion of the point is non-uniform, the above equation
does not give the speed of the point at each instant in the interval,
but gives only the average speed for the time interval, M. The
instantaneous speed is the average speed over an indefinitely small
period of time including the instant, or, expressed in mathematical
form, the speed at any instant is,
- . ..As ds .. .
v= Limit -r~=-T- . (1)
A^^O ^^ ^^
The direction of v, as already noted, is tangent to the path at the
point on the path where the moving particle is located at the
instant.
The unit of velocity may be any convenient unit of length
per unit of time; such as, foot per second (ft./sec), mile per hour
(mi./hr.), centimeter per second (cm./sec), etc.
In order to find the value of v by differentiation, as indicated in
equation (1), s nmst be expressed algebraically in terms of I.
ILLUSTRATIVE PROBLEM
267. A point moves along a straight line path according to the law,
s = 6<2-f-4. What is the velocity of the point at the end of 5 sec?
Solution.— v==j^ = j^ (6f2+4) =12«.
Hence, when ^ = 5 sec, the velocity of the point is,
y = 12X5 = 60 ft./sec
228
MOTION OF A PARTICLE
PROBLEMS
268. A point moves along a path according to the law s = St^+2t+t-'^,
s and t being in feet and seconds, respectively. What is the speed of the point
when t = 2 sec? Ans. y = 13.75 ft./sec.
269. A point moves along a straight line according to the law v = t^-\-4t^-\-2.
If s =4 ft. when t = 2 sec, what is the value of s when / = 3 sec?
Ans. s =47.58 ft.
113. Distance-time Graph. — If a point moves in such a way
that the distance s, in equation (1) of Art. 112, cannot be ex-
pressed algebraicall}' in terms of t, or, if the relation between s and
t leads to a complex equation, the speed maj^ often be found con-
veniently, by a graphical process, from the distance-time (s-t) curve.
-250
/
-200
(s-t) graph N^
/
/
c
-150
>
/.
^^
-100
.^
1
B
-50^
/^
1,0
1,5 2!o 25
30
35 4
'
t, seconds
Fig. 272.
Thus, instead of expressing the relation between s and t algebra-
ically, it is shown graphically by plotting a series of points, the
rectangular coordinates of each point being simultaneously values
of s and t. One such graph is shown in Fig. 272.
. ds .
Since the slope to the {s-t) curve is represented by -r. and since
ds
v=-r. (Art. 112), then the slope at any point of the (s-t) graph
represents, to some scale, the speed at the corresponding
instant.
ANGULAR VELOCITY _ 229
ILLUSTRATIVE PROBLEM
270. If the curve in Fig. 272 is the s-t graph for a point moving on a
straight-line path, what is the velocity of the point at the end of 20 sec?
Solviion:
c,, ^ . BC 2 units verticallv 1
Slope at A
AB 4 units horizontally 2*
But 1 unit vertically represents 50 ft. and 1 unit horizontally represents 5 sec.
Hence,
Unit slope = ~ '- = 10 ft. /sec.
5 sec.
Therefore velocity at end of 20 sec. =^X10
= 5 ft. /sec.
114. Angixlar Velocity. — The angular velocity of a moving
particle is defined as the time rate of angular displacement of the
particle.
If equal angular displacements occur in equal time mtervals,
however small, the motion is said to be uniform, and the angular
velocity, co, is expressed as the ratio of any angular displacement,
A^, to the time interval, A^, during which the displacement occurs.
Thus,
A^
At
If unequal angular displacements occur in equal time intervals,
the motion is said to be non-uniform or variable. For such motion
the above equation gives the average angular velocity during the
time interval A^. When the angular velocity varies during the
interval, its value at any instant is the average velocity over an
indefinitely small time interval including the instant. Or,
expressed mathematically, the instantaneous angular velocity is,
<D = Limit -J- = -r:.
A^=0^^ ^'
The unit of angular velocity is any coiivenient unit of angular
displacement per unit of time; such as, degree per second
(deg /sec), revolution per minute (r.p.m.), radian per second
(rad./sec), etc.
230
MOTION OF A PARTICLE
In order to determine the angular velocity from the above
equation, 6 must be expressed algebraically in terms of t, a require-
ds
ment similar to that met in using the equation z; = -7- in Art. 112.
The angular velocity may also be found from an angular displace-
ment-time {6-t) graph by a graphical method similar to that used
in determining the hnear velocity in Art. 113.
115. Relation between Linear and Angular Velocities. — In
many problems in kinematics it is convenient to express the
linear velocity of a point in terms of its angular velocity. The
relation between the two velocities may be found as follows:
Let a point M move on a circular path of radius r (Fig. 273a);
Fig. 273.
let V be the Hnear velocity of the point at any instant and let co
be the angular velocity of the point at the same instant. From
definition,
ds , dd
v = -T. and 0) = -^-.
dt dt
But the distance ds traversed in the time dt may be expressed in
terms of the corresponding angular displacement dd, by the
equation ds = rdd (Art. 111). Therefore,
rdQ
V = — = Tea.
dt
Hence, at any instant, the Unear velocity of a point moving on a
COMPONENTS OF VELOCITY
231
circular path is the product of its angular velocity ^ (with respect
to the center of the path) and the radius of the path.
If the particle does not move on a circular path, the equa-
tion v=oir is also true if r is the radius of curvature of the
path at the given position of the particle, and if co is the
angular velocity of the particle with respect to the center of curva-
ture as the pole.
Further, if the point moves on a curve of any form and the
center of curvature is not taken as the pole (Fig. 2736), then the
term pw, where p denotes the radius vector to the point, gives one
component only of the linear velocity, as is shown in the next
article.
PROBLEMS
271. A flywheel 6 ft. in diameter rotates at 120 r.p.m. Find the hnear
velocity, in feet per second, of a point on its circumference.
272. A rod 4 ft. long rotates in a horizontal plane about a vertical axis
through one end of the rod, so that the
linear velocity of its mid-point is 60 ft. per
sec. Find the angular velocity of the rod,
in r.p.m.
273. An automobUe is traveling at 30
miles per hour on a straight road. An
observer is stationed at 0 (Fig. 274). Find
the angular velocity of the automobile with
respect to the observer (a) when the auto-
mobile is at A; (6) when at B.
116. Components of Velocity. — It
is frequently convenient to find the
velocity of a moving point by de- pj^ 274.
termining its components, or to deal
with the components of the velocity instead of the total velocity.
The assumption that velocity may be resolved into components
1 When angular velocity is involved in the same expression with linear
velocity, the angular velocity is expressed in radians per unit of time. Since
the number of radians turned through is a length along a circular arc divided
by the length of the radius and, hence, is a ratio of length to length, it has no
dimensions. Thus the dimensional equation for y = cor is ^= (yX-^) Xl/,
and hence the equation is dimensionally correct (Art. 19). Likewise, angular
displacement and angular acceleration are expressed in radians and radians
per second^, respectively, under similar conditions.
232
MOTION OF A PARTICLE
follows from the fact that velocity is a vector quantity. Its truth,
however, is abundantly confirmed by the agreement of results,
deduced by the use of velocity components, with observed facts.
Two sets of components only are here determined; namely, the
axial components (vx and
Vp), parallel, respectively, to
the X- suid-y axes, and the
radial and transverse com-
ponents (vr and Vt), par-
allel and perpendicular,
respectively, to the radius
vector (Fig. 275).
Thus, since the linear
velocity of a point in any
direction is the rate at
which the position of the
point is changing
given direction, the axial components of the velocity v
moving point M (Fig. 275) are given by the expressions,^
Fig. 275.
m
of
the
the
Vx =
dx
dt
and Vy =
dy
df
ov. if the coordinates, x and y, of the particle change uniformly,
then,
Aa: X2 — X1
Vx = ^-r =
and Vv = ~-
^y 2/2-2/1
A^ t2-ti " At t2-ti
Likewise, at any instant, the transverse and radial components,
Vt and Vr, are expressed as follows:
d?
Vt = -3T and
at
Vr =
dV
''■ These equations follow from the definition of velocity provided that the
assumption is made that the components of a displacement may be treated,
in all respects, like the total displacement itself and that the vector sum of the
rates of change of the components is the same as the rate of change of the total
displacement. The equations may be derived through an algebraic series
of steps beginning with the definition of a velocity component, Vx = v cos a
ds
and substituting therein the values of v and cos a] namely, v=-t and cos a
dt
dx
-J-. This method gives an apparent definiteness to the conclusion but is
ds
lacking somewhat in its emphasis on the fundamental characteristics of the
quantities involved.
COMPONENTS OF VELOCITY
233
in which pdO and dp are the component displacements in the
transverse and radial directions during the time interval dt, as
shown in the preceding article, and co is the angular velocity of the
particle with reference to the pole 0. It will be noted that when 0
is chosen as the center of curvature of the path the transverse
component of the velocity becomes the total velocity, tangent to
the path, Vr then being equal to zero.
Since the pole 0 is arbitrarily chosen, the components Vq'
and Vr are different for different positions of the pole or
origin, whereas Vx and Vy are independent of the origin but
depend on the directions of the coordinate axes. Further, if
the path of the point is circular and the pole 0 is at the center or,
in general, if 0 is the center of curvature of the path, at th-e instant,
then dp and -t- are zero and the expressions for Vt and v are the
same.
ILLUSTRATIVE PROBLEM
274. In the quick-return mechanism shown in Fig. 276, 00i = 18 in. and
the crank OA =8 in. If the angular velocity of the crank is 40 r.p.m., what
is the velocity of the block A? Find graphic-
ally the component of the velocity of A
perpendicular to the rocker arm OiM (the
transverse component). Find also the angu-
lar velocity of OiM.
Solution. — The block moves on the cir-
cular path of radius r = 8 iq. Let co denote
the angular velocity of A with reference to
the pole 0 and coi the angular velocity of A
with reference to the pole Oi (that is, let coi
denote the angular velocity of OiA, or p).
The direction of the velocity of A is tangent
to the circle and its magnitude is,
t; = a;r=^5^X^ = 2.79 ft./sec.
By resolving v, graphically, into its transverse
1 in.= 4ft./sec
Fig. 276.
and radial components as shown in the figure, the following values are found.
But,
Vr = 1 . 8 ft./sec. and vr=2.17 ft./s
By measuring, OiA is found to be 11.7 in.
234
Therefore,
MOTION OF A PARTICLE
ui = '■,., ^ = 1 . 84 rad./sec. = 17.6 r.p.m.
11.7
PROBLEMS
275. A rigid body composed of two disks of different diameters (Fig. 277)
turns about the center 0 so that a point on the circumference of each disk
Fig. 277.
Fig. 278.
moves with a constant angular velocity of 80 r.p.m. If n = 2 ft. and ra = 3 ft.,
what is the velocity of body A and of body B? A unwinds from the large disk
as B winds up on the small disk. How far will A and B travel in 6 sec?
r^
Fig. 279.
276. The link BC (Fig. 278)
is 2 ft. long and has an angular
velocity, on, of 50 r.p.m. If the
link BA is perpendicular to BC,
at the instant considered, what
is the angular velocity of the
crank OA if OA is 9 in. long?
277. If the angular velocity
of the oscillating arm OM (Fig.
279) is 40 r.p.m., in the position
shown (0 = 30°), find the trans-
verse component of the velocity
of the block A, referred to 0 as
the pole. Also find the total
velocity of A .
Ans. fA =9.3 ft. /sec.
278. The cam shown in Fig. 280 revolves about the axis 0, causing the
roller A to change its x-coordinate at the rate of 4 in. per second when 0 = 30°.
Find the angular velocity of the bell-crank AOiB if OiA is 18 in.
Ans. (i)«=4.24 r.p.m.
LINEAR ACCELERATION
m
Fig. 280.
117. Linear Acceleration. — ^The acceleration of a moving point,
at any instant, is the time rate at which its velocity is changing at
the instant. The 'velocity of the point has, at any instant, a
definite magnitude (speed) and a definite direction. A change in
the velocity occurs if either its magnitude or its direction changes.
Hence the acceleration of the point may be the rate of change of
the velocity due to a change in the magnitude of the velocity, or it
may be the rate of change of the velocity due to a change in the
direction of the velocity. These statements are not mere arbi-
trary statements; they are in accord with our experience, for
experience teaches that a particle tends to maintain the direction
of its velocity as persistently as it tends to maintain the magnitude
of its velocity. The two properties of velocity are inherent,
independent properties. Hence,
Whenever the velocity of a particle changes in magni-
tude the particle is accelerated.
Whenever the velocity of a particle changes in direction
the particle is accelerated.
If the direction and magnitude of the velocity change simul-
taneously, the acceleration of the particle is then the vector sum of
the two accelerations which arise as the result of the separate
changes.
In rectilinear motion the velocity of a particle can change in
magnitude only, whereas in curvilinear motion the velocity of the
236
MOTION OF A PARTICLE
particle may change in magnitude but must change in direction.
Therefore a particle cannot have a curvilinear motion without
being accelerated.
PROBLEMS
279. A quarter-mile track is made up of two straight parallel sides con-
nected at the ends by semicircles. A boy
runs the quarter mile in 50 seconds at uni-
form speed. Does the boy have an accelera-
tion while on the straight sides? While on
the curved ends? Why?
280. A particle moves with a constant
speed of 10 ft. per sec. on the path shown
in Fig. 281. Does the particle have an
acceleration when at A? at 5? at C?
at D? What is the value of the accelera-
tion of the particle when at A, and at D?
Fig. 281.
118. Tangential and Normal Acceleration. Tangential Accel-
eration.— The facts stated in the preceding article make it possible
to determine, for a given motion, whether or not a particle is being
accelerated. The exact magnitude and direction of the accelera-
tion produced in each of the two ways mentioned in the preceding
article will now be found.
In Fig. 282 let a point move so that its velocity changes in
^1
Path
'ȣ
^2
I?!
Av^
Fig.
282.
magnitude only. That is, let the point move on a straight line
path. Let vi be the velocity at one instant and let V2 be the
velocity after the time interval A^. If the velocity changes
uniformly, the magnitude of the acceleration is the ratio of any
change in the velocity, Av, to the time interval, A^, during which
the change Av occurs. Since V2 and vi have the same direction,
Av is the algebraic difference of V2 and vi as well as the vector
difference. Hence, the magnitude of the acceleration for uni-
formly accelerated, rectilinear motion is,
Av V2—V1
At t2-ti
(1)
TANGENTIAL AND NORMAL ACCELERATION 237
If the velocity of the moving point in Fig. 282 does not change
uniformly, then equation (1) gives only the average acceleration
during the period A^. When the acceleration varies from instant
to instant its value at any instant is the average acceleration during
a very small time interval including the instant. Or, expressed
mathematically, the instantaneous acceleration for rectilinear
"^«^i«^i«> ,. .A. dv
a= Limit -- = -T., (2)
At=0 ^^ ^^
V being the velocity of the particle at the instant. In order to
find a from the above equation, v must be expressed algebra-
ically in terms of t.
The direction of the acceleration of the particle is the same as
that of the change of velocity, A^;, but Av is parallel to v, that is,
along (or tangent to) the path. If V2 is smaller than vi, the sense
of Av is negative, that is, opposite to that of v, and hence
the acceleration then is negative.
If the particle moves on a circular path so that, as the
velocity changes from vi to V2f its direction as well as its mag-
dv
nitude changes, then -^ expresses only that part of the accelera-
tion which is due to the change of the magnitude of the velocity.
dv
And the direction of -r. at any instant, as already noted, is parallel
to the velocity, or tangent to the path at the point where the par-
ticle is located at the instant. '
gential acceleration and is written,
dv
dv
Hence, -r is the total acceleration of a moving point, only, when
the point has a rectiUnear motion.
Normal Acceleration. — The acceleration produced by a change
in the direction of the velocity of a point will now be considered.
In Fig. 283 let a particle move on a circular path with constant
speed. That is, let the velocity change in direction only. At one
instant the velocity is vi, and during a time interval A^ the
velocity changes to V2. The change in the velocity during the
time interval, that is, the vector difference between V2 and vi, is
dv
tide is located at the instant. Therefore, -r: is called the tan-
dt
238
MOTION OF A PARTICLE
AT=2Usin-
Av, as shown in Fig. 283, since Av is the velocity that must be
added to vi to give V2, or, V2 is the resultant of vi and Av. For
motion in which the direction
of the velocity does not
change (rectilinear motion),
the vector difference of the
velocities of the point and
the scalar (or algebraic) dif-
ference are the same, but in
curvilinear motion (Fig. 283)
the algebraic difference is
zero, whereas the vector dif-
ference, or change, of the
velocities, is Av, the mag-
nitude of which is,
A2; = i;2
Vl
o • ^<9 _ .AS
= 22^1 sm Y = 2z;2 sm —
= 2v sin
A^
2'
in which v is the magnitude of the velocity at any instant,
since z;2 = 2^1 = 2^. Hence, since the acceleration at any instant
is the average acceleration during an indefinitely small time interval
including the instant, the magnitude of the instantaneous ac-
celeration due to the change of direction of the velocity is.
2v sin
a = Limit
At^O
Ad
2'
At
Now, as A^ approaches zero, sin -^ approaches — and — approaches
Z Z At
-77. Hence, in the limit,
at
dd
dd
But -r. is the angular velocity of the moving point which corre-
TANGENTIAL AND NORMAL ACCELERATION 239
spends, at any instant, to the linear velocity v. Further, t; = cor
(Art. 115). Therefore,
r
The particular form most convenient for use depends upon the
particular problem under consideration.
The direction of this acceleration is the limiting direction of
^v as M approaches zero, and since A^ approaches zero as M
approaches zero, it will be seen from Fig. 283 that the hmiting
direction of Lv is perpendicular to vi and, hence, is toward the
center about which the point is turning, that is, normal to the
path. This acceleration is, therefore, called the normal accelera-
tion and is written,
If the path on which the particle moves is not a circular path, the
above expression also holds provided that r is the radius of curva-
ture of the path at the point where the particle is located at the
instant, and that co is the angular velocity with respect to the cen-
ter of curvature as a pole. Summarizing, the two following
important theorems may be stated:
I. When the velocity, v, of a particle changes in mag-
nitude, an acceleration is produced the value of which is —r\
at
its direction at any instant is parallel to that of the velocity,
that is, tangent to the path at the point at which the
particle is located at the instant. Thus,
dv
IL When the velocity, v, of a particle changes in direc-
tion, an acceleration is produced the value of which is
vo3 = oi^r = — ; its direction, at any instant, is perpen-
dicular to the velocity, towards the center of curvature
of the path, at the point where the particle is located at
the instant. Thus,
2 ^
a„ = i;co = co^r = — .
r
240
MOTION OF A PARTICLE
The unit of acceleration is any convenient unit of velocity
per unit of time; such as, foot per second per second (ft./sec.^),
mile per hour per second (mi./hr./sec), etc. ^
ILLUSTRATIVE PROBLEM
281. Two pulleys are connected so that they turn together about the center
0 (Fig. 284), causing the weight A to unwind and the weight B to wind up.
If the angular velocity of the points M and P change uniformly from 10 r.p.m.
to 60 r.p.m. during a period of 2 sec, find: (a) the tangential acceleration of
each of the two points at any instant during the 2 sec; (6) the acceleration of
A and of B; (c) the total acceleration of M at the beginning and of P at the
end of the two-second period.
Solution. — (a) The velocity of the point M changes in magnitude from
3
^TT ft. /sec in 2 sec. Hence the
Vi = <airi= — :p^ — Xt = 7 ft. /sec to v-^ = o:-2ri
60
4 4
tangential acceleration of M is,
(aM)t =
V2—V1
At
2
57r = 1.96ft./sec.2.
The velocity of P changes from ^ ft. /sec to
o
2ir ft. /sec in 2 sec. Hence,
Mr
V2— «i 2ir — lT 5
At
x = 2.62 ft./sec.2
(b) The magnitude of the velocity of any
point on the circumference of the small pulley
changes the same amount as does the velocity
of A. Therefore, the tangential acceleration of
M is the same in magnitude as the acceleration
of A. Hence,
0/i = (aM)i = 1.96 ft./sec^
a5 = (ap)^ = 2.62ft./sec.2.
(c) The normal acceleration of M at the beginning of the 2-sec. period is,
(aM\ = c.M= (^^^) 'x^ = 0.82 ft./sec^
and the normal acceleration of P at the end of 2 sec is,
(ap)„ = o„V.= (55|^) 'xi|=39.4 ft./sec.^
Therefore the total acceleration of M is,
aAf = V(1.96)2+(.82)2 = 2.12 ft./sec^
SPEED-TIME GRAPH
And the total acceleration of P is,
241
ap = V(2.62)2 + (39.4)2 = 39.5 ft./sec.2.
The acceleration of M, only, is shown in Fig. 284.
PROBLEM
282. A train, while rounding a curve of \ mile radius, changes its speed
uniformly from 20 mi./hr. to 30 mi./hr. in 20 sec. What is the total accelera-
tion of the train at the beginning and at the end of the 20-sec. period?
119. Speed-time Graph. — In finding the total acceleration
dv
of a point from the equation o,=--r. (in rectiUnear motion), or,
in finding the tangential acceleration from the equation at = -r
dii
(in curvilinear motion), by the calculus method, v must be expressed
algebraically in terms of t. The relation between v and t may some-
times be shown more conveniently, graphically, by plotting a
speed-time {v-t) graph the coordinates of any point on which are
simultaneous values of the speed and time of the moving point
(similar to the distance-time graph in Art. 113). One such graph
is shown in Fig. 285. Since the slope of the speed-time curve is
dv
represented by -r, the slope of the v-t graph at any point repre-
sents, to some scale, the acceleration of the moving point at the
corresponding instant. It is important to note that this method
gives only that acceleration which is due to the change in the
magnitude of the velocity. It is applicable, therefore, only to the
total acceleration for rectihnear motion or to the tangential accel-
eration for curvilinear motion.
ILLUSTRATIVE PROBLEM
283. In Fig. 285 is represented
the speed-time graph for a point
which moves on a circular path.
What is the magnitude of the tan-
gential acceleration at the end of
15 sec?
Solution. — at at the end of 15 sec.
= slope at A
=^^ = 2 ../sec.
100-
40--
it-^
15 20 25
t, sec.
FiQ. 285.
H H
30 35
242
MOTION OF A PARTICLE
PROBLEMS FOR ARTICLES 118 AND 119
284. A train running on a straight track has a speed of 30 mi. per hr. when
passing a certain mile post and a speed of 50 mi, per hr. when passing the
next mile post. If the rate of change of speed is uniform and the time required
to travel the mile is 1.5 min., what is the acceleration of the train?
285. A point moves on a circular path having a radius of 2 ft. If its speed
changes imiformly from 100 ft. per min. to 240 ft. per min. during a period
of 3 sec, what are the tangential and normal accelerations (a) at the begin-
ning of the period; (6) at the end of the period?
286. A flywheel 8 ft. in diameter turns so that the angular velocity of a
point on its rim changes from 100 r.p.m. to 40 r.p.m. at a uniform rate during a
period of 4 sec. Find the tangential acceleration of a point on the rim during
the 4 -sec. period. Find the total acceleration of a point on the rim at the
end of the period. Ans. 0^ = 6.28 ft, /sec. 2; a = 70.5 ft. /sec. ^
Fig. 286.
287. In Fig. 286 is shown the speed-displacement graph for the crosshead,
C, of a steam engine. The crank OA revolves at a constant speed of 233
r.p.m. Show that the slope to the speed-displacement curve represents the
ratio of the acceleration of C at any instant to its speed at the same instant.
If the scale of abscissas is 1 in. = 1.75 ft. and if BC measures 0.33 in.,
what is the acceleration of the crosshead in the position shown? How could
the position of C be found for which C has zero acceleration?
120. Angular Acceleration. — The angular acceleration of a
moving point is the time rate of change of the angular velocity of
the point. If the angular velocity, co, of the point changes
uniformly, the angular acceleration, a, is expressed by the ratio of
any change, Aco, in the angular velocity to the corresponding time
interval, A^. Thus,
Aco 0)2 — wi
^^At^ t2-h '
LINEAR AND ANGULAR ACCELERATIONS 243
If the angular velocity of the point does not cha,nge uniformly,
the above equation gives only the average acceleration during the
time interval A^. When the acceleration varies from instant to
instant its value at any instant is the average acceleration during
an indefinitely small time interval including the instant. Or,
expressed mathematically, the instantaneous angular acceleration
is,
_ . .^ Aco dco
a= Limit -jr,=~rr^
And since co = -7-, a may also be expressed by the equation,
at
"" dt\dt)
df
In order to find a from the above equations, to and 6 must be
expressed in terms of t.
The unit of angular acceleration is any convenient unit of
angular velocity per unit of time; such as, degree per second per
second (deg./sec.^), revolution per minute per second (rev./min./
sec.) radian per second per second (rad./sec.^), etc.
PROBLEMS
288. What is the angular acceleration of a point on the rim of the fljrwheel
having the motion described in Problem 286?
289. What is the angular acceleration of the train having the motion
described in Problem 282? Ans. a = 0.000555 rad./sec.^
290. A particle moves on a circular path according to the law d=3t^-{-2t.
What is the angular velocity and the angular acceleration of the particle at
the end of 4 sec?
121. Relation between Linear
and Angular Accelerations. — In many
problems in kinematics it is con-
venient to make use of the relation
between the linear and angular ac-
celerations of a point. This relation
may be found as follows: In Fig. 287
let a point move on the circular
path shown, 0 being the center of the Fia. 287.
244
MOTION OF A PARTICLE
path and r the radius. If the magnitude of the velocity of the
point changes (assumed to increase) there is, at any instant during
the change, a tangential acceleration given by the equation,
a,=
dv
di'
But,
v = roi.
d{ojr)
dw
But -J- is the
at
Therefore q<= \"^=r^, since r is constant.
at at
angular acceleration, a, of the point at the given instant. Hence,
Therefore the tangential acceleration,^ at any instant, of a
point moving on a circular path is equal to the angular accelera-
tion of the point, at the same instant, about the center of the path
Fig. 288.
FiQ. 289.
times the radius of the circle. If the particle does not move on a
circular path, the equation is also true provided that r is
the radius of curvature of the path at the given position of the
particle and that a is the angular acceleration of the particle
with reference to the center of curvature.
If, however, the point moves on a path of any form and the
pole is not taken at the center of curvature of the path at the given
position of the particle (Figs. 288 and 299), then, although the
equation 0,=-^ is true, a, is not equal to pa since v is not equal to
p« (Art. 115).
* See footnote for Art. 115.
AXIAL COMPONENTS OF ACCELERATION
245
The normal acceleration, rco^, of the particle (not shown in
the above diagrams), unlike the tangential acceleration, is inde-
pendent of the angular acceleration. It depends on the angular
velocity at the instant, and not on the rate at which the angular
velocity is changing at the instant.
PROBLEMS
291. As the drum (Fig. 290) turns, the weight A is wound up with decreas-
ing speed. If its speed decreases 20 ft. /sec. each
second, what is the angular acceleration of a
point on the rim of the drum?
292. A rod 3 ft. long is rotated in a hori-
zontal plane about a vertical axis through one
end of the rod so that the angular velocity in-
creases uniformly from 10 to 40 r.p.m. in 3 sec.
Find the tangential acceleration of the mid-point
of the rod. Ans. o< = 1.57 ft. /sec'
122. Axial Components of Accelera-
tion.— The component, in a given direc-
tion, of an acceleration of a point is the Fig. 290.
rate of change, in the given direction,
of the velocity of the point. In general, the rate of change, in
the given direction, of the velocity of the point is due partly to
a change in the magnitude and partly to a change in the direction
of the velocity. It will be noted
that the tangential and normal ac-
celerations, at and an, which are dis-
cussed in the preceding articles are
components of an acceleration, one
of which, at, is the rate of change of
the velocity due to a change in its
magnitude only, and the other, an,
is the rate of change of the velocity
due to a change in its direction
only. Thus, if a point moves on a
curved path with increasing (or decreasing) speed (Fig. 291),
both the magnitude and the direction of the velocity changes
and the expressions for the total acceleration as found in the
preceding articles are,
Fig. 291.
246
MOTION OF A PARTICLE
dv jp o
dt r
and the direction of a is expressed by the equation,
tan 0 =
(h
Acceleration, like any directed quantity, may be resolved into
various sets of components. The axial components, in addition
to the tangential and normal components, are frequently desired.
The axial components {ax and ay) are parallel, respectively, to the
X- and 2/-axes. Expressions for the axial components may be found
as follows :
If the velocity of a point is resolved into components in arbi-
FiG. 292.
trarily selected directions, as, for example, parallel to the x- and y-
axes, then the acceleration of the point is the (vector) sum of the
rates of change in these components. But since the directions
of the velocity components are fixed, the components can change
in magnitude only, and hence the acceleration in each of the
dv
selected directions is of the type -5- (Theorem I, Art. 118). There-
fore the axial components of the acceleration of a point (Fig. 292)
are,
dVx , dvy
a. = -^ and a,^^
AXIAL COMPONENTS OF ACCELERATION 247
dx dti
And since Vx = -Tr and Vy = -^, the axial components may also be
written,
In order to determine the axial components from the above equa-
tions, Vx (or x) and Vy (or y) must be expressed in terms of t, unless
Vx and Vy change uniformly in which case,
_AVx_v''x—v'x
^'~ At " t2-ti '
and,
^^Jy V"y-V'y
Qv =
At t2-ti
It is important to note that the velocity of a point in a given
direction may be zero without making the acceleration in that
direction zero. For example, a point moving on a circular path
has no velocity component normal to the path, that is, Vt is the
total velocity {vn = 0) but there is a normal acceleration (rate of
change, in the normal direction, of the velocity) the magnitude of
which as already shown, is —. Similarly, a ball thrown horizontally
from a window has no vertical velocity just as it leaves the window,
that is, Vy = 0, but the acceleration of the ball at that instant is
ay = S2.2 ft./sec.2 In other words, Vy is changing through its zero
value at the rate of 32.2 ft, /sec. each second. Likewise, the velocity
of the crosshead of a steam engine is zero at the end of the stroke,
dv
but its acceleration, -j, has a large value as the velocity changes
through its zero value.
PROBLEMS
293. The total acceleration of a point on the rim of a pulley, at a given
instant, is 40 ft./sec.^ in a direction making an angle of 20° with the radius to
the point. If the radius of the pulley is 18 in. what is the angular velocity
and the. angular acceleration of the point at the given instant?
Ans. (0 = 5.01 rad./sec; a = 9. 12 rad./sec.^
294. A particle moves on the path xy = 4:, according to the law, x = 2t^.
(a) Find the x- and ^/-components of the velocity and of the acceleration at the
end of 2 sec. (h) Determine the total acceleration by combining the com-
ponents graphically, (c) Find the tangential and normal components graph-
ically by resolving the total acceleration in the tangential and normal direc-
tions.
248 MOTION OF A PARTICLE
123. Motion of a Projectile. — The equations developed in the
preceding article will here be used in the analysis of the motion of a
projectile. The actual motion of a projectile is influenced by a
number of factors, such as the rotation of the projectile due to the
rifling of the barrel, wind velocity, humidity, etc., which require
modifications in the results found from the assumed ideal condi-
tions. The motion of a projectile moving without rotation in a
vacuum will here be considered. The motion may be studied by
treating the components of the motion in the x- and ^/-directions
as follows:
Let a projectile be given an initial velocity w in a direction
making an angle 6 with the hori-
zontal and let the point of pro-
jection be taken as the origin as
shown in Fig. 293. The accelera-
tion of the projectile during its
flight is constant and is directed
U r ^ vertically downward. This con-
Pjq 293 stant acceleration, denoted by gr,
is the acceleration due to gravity,
since the projectile is a freely falling body, that is, no forces
except the earth-pull (its weight) act on it. Hence,
' '=0 or Vx=Ux=u cos 0, (1)
and,
Vv — U
ay= =—g or Vy=Uy — gt = u sin d—gty . . (2)
considering v„ and Uy positive when directed upward.
The horizontal distance, x, traveled in any time interval t is,
X = Vx't = U COS d't, ...... (3)
and the vertical distance, y, traveled in any time interval t is the
average velocity times the time interval. Hence,
y^rimI + (^iml^M(l.t^u sin e-t-igfi. . . (4)
Particular values found from the above equations are of special
importance, namely, the range on a horizontal plane, the greatest
height, and the time corresponding to each of these distances.
MOTION OF A PROJECTILE 249
Time of Flight. — Since y equals zero when the projectile reaches
the a;-axis, the time of flight, U, is given by t in the equation,
y = u mi d-t — \ gi^ = ().
Therefore,
«,=??^ (5)
Range. — ^The range, r, equals the value of x in equation (3)
when t = tr. Therefore,
u^ sin 26
r=
9
Time to Reach Greatest Height. — When the projectile reaches its
greatest height, v^^^O. Hence, the time, //,, required for the pro-
jectile to reach its greatest height is given by t in the equation,
Vy = u sin d—gt = 0.
Therefore,
1^ sin ^
h — —.
Greatest Height. — The greatest height, h, will be given by y in
equation (4) when t has the value th. Hence,
The equation of the path of the projectile (called the trajectory)
may be obtained by eliminating t from equations (3) and (4),
which gives the following equation :
ox
2u^ cos^ 0
Hence the trajectory is a portion of a parabola with its axis vertical.
PROBLEMS
295. A bullet is projected upward at an angle of 60° with the horizontal
with a velocity of 2000 ft. /sec. Find the range and time of flight.
Ans. r = 108,000 ft.
296. A shot is fired from a gun on the top of a cliff 400 ft. high, with a
velocity of 768 ft. /sec, the angle of elevation of the gun being 30°. Find
the range on a horizontal plane through the base of the cliff. Ans. r = 5543 yd.
250 MOTION OF A PARTICLE
124. Uniformly Accelerated Rectilinear Motion. — Many
examples of straight-line motion with constant acceleration occur
in engineering practice, such as the motion of a freely falling
body or of a train leaving a station imder the action of a con-
stant draw-bar pull. Further, from the preceding article it will
be noted that one of the components of a curvilinear motion
may be a uniformly accelerated rectilinear motion. The rela-
tions between the distance, time, velocity, and acceleration, for
imiformly accelerated rectilinear motion may be deduced as
follows :
By definition,
Av v—u ...
"=Ar-r (^>
or, v = u-\-at, (2)
in which u and v are the initial and final velocities, respectively,
corresponding to the time interval A^ or simply t.
Since the velocity increases or decreases uniformly, the average
velocity is —^ and the distance, s, traveled in time t is,
-^•* ....... (3)
_u-\-u-\-at
~ 2 '^
=ut+iat^. ...'... (4)
By eliminating t from equations (2) and (4), the following equa-
tion is obtained :
iP = u^-\-2as (5)
Hence, for uniformly accelerated, rectilinear motion, the fol-
lowing equations express the relations between the distance, time,
velocity, and acceleration,
v=u+at, (6)
s=ut+iat^, (7)
v^ = u^-\-2as, (8)
-T^' (9)
UNIFORMLY ACCELEIL\TED RECnLINK\R MOTION 251
If the point starts from rest, u in each of the above equations
is zero, and the equations reduce to simpler forms. For freely
falling bodies, a is the acceleration due to the earthpull on the
body and is usually denoted by g, its value being 32.2 ft. sec.^
(approximately). Further, the above equations also applj- to
the motion of a point moving on a curved path if the speed
changes imiformly and if the tangential acceleration Ot of the
point is used for a.
PROBLEMS
297. Deduce equations (6) and (7) by calculus methods, starting with the
ds
dv
equations, v = --r. and o = ;ii
298. Draw a speed-time graph for uniformly accelerated, rectilinear motion
and deduce equations (7) and (9) .
299. The brakes are set on a train running at 30 mi. per hr., when ^ mi.
from a station. The train slows down uniformly, coming to rest at the station.
Find the acceleration, and the time in stopping.
Ans. a =0.306 tt./aec.K < = 120sec.
300. If the maximum allowable
speed of an elevator is 800 ft./min. and
if it acquires this speed uniformly in a
distance of 12 ft., what acceleration
does it have?
301. A train in starting is uniformly
accelerated and attains a speed of 60
mi./hr. in 5 min. After running for a
certain period of time at this speed, the
brakes are appUed and it stops at a
uniform rate in 4 min. If the total
distance traveled is 10 mi. find the total
time. Ans. f = 14.5 min. „
Fig. 294.
302. If the cam A (Fig. 294) moves
to the left, changing its velocity imiformly 5 in. per sec. each second, what
is the acceleration of the rod B (a) before the pin comes in contact with the
cam; (6) just as it comes in contact with the cam; (c) while it is in contact
with the cam?
125. Uniformly Accelerated Circular Motion. — Many prob-
lems involving the motion of a point on a circular path with con-
stant angular acceleration occur in engineering practice. The
relation between the angular displacement, angular velocity.
252 MOTION OF A PARTICLE
angular acceleration, and time may be deduced in a manner similar
to that used in the preceding article.
By definition,
ACO CO — coo /1X
or, (jj = o3o-\-at, (2)
in which coo and co are the initial and final angular velocities,
respectively, corresponding to the time interval A^ or simply t.
The angular displacement A^, or simply 6, is the average angu-
lar velocity times the time interval. Hence,
e='^-t (3)
_coo+^o+Q'^ .
~ 2 '^
= mt-{-iat^ (4)
By eliminating t from equations (2) and (4) , the following equation
is obtained,
Co2=COo2 + 2c^^. (5)
To uniformly accelerated circular motion, therefore, the fol-
lowing equations apply :
<ji = (jio-\-at, (6)
Q = mt+hott^ (7)
a)2 = (oo2+2Q{e, (8)
e=^./ (9)
It will be noted that the above equations may be derived from
the corresponding set of equations in Art. 124 by making use of
the equations, w = coor, v=o)r, at = ra, and s = rd.
PROBLEMS
303. Derive equations (6) and (7) by calculus methods starting with the
dd , do}
equations, '«'=,. and oc = ^.
304. A wheel starting from rest turns so that a point on its rim has its
angular velocity increased uniformly to 200 r.p.m. in 6 sec. After turning for a
SIMPLE HARMONIC MOTION
253
certain period of time at this speed a brake is applied and the wheel stops at a
uniform rate in 5 sec. If the total number of revolutions is 3100, find the
total time.
126. Simple Harmonic Motion. — If the velocity of a point does
not vary uniformly, the acceleration is not constant, and hence
the equations of Art. 124 do not apply. One special case of
rectilinear motion with variable acceleration is simple harmonic
motion. A simple harmonic motion is defined as the motion of a
point in a straight hne such that the acceleration of the point is
proportional to the distance, x, of the point from some fixed
origin 0, in the line and is directed toward 0. Or expressed
mathematically,
«=-5^=-^^^ • (1)
where A; is a constant and the negative sign indicates that the sense
of the acceleration is opposite to that of the displacement x (Fig.
295), that is, a is negative when x is positive, and positive when
X is negative.
One example of a simple harmonic motion is the motion of a
weight attached to the lower end of
an elastic spring (the upper end
being fixed) which is allowed to
vibrate freely. The motion of the
crosshead of a steam engine closely
approximates a harmonic motion,
the approximation becoming closer ^^
0
a--kx
-t
1
1
Fig. 295.
Fig. 296.
as the ratio of the length of the connecting rod to that of the
crank increases. The motion of an oscillating pendulum also
approximates closely a simple harmonic motion if the arc through
which the pendulum swings is small. Further, if a point moves
with constant speed in a circular path, the motion of the projection
of the point on a diameter of the circle is a simple harmonic
motion. This last statement may be proved as follows: In Fig.
254 MOTION OF A PARTICLE
296 let ilf be a point moving with constant speed, vo, and
angular velocity, co, along a circle of radius r. If t is the time
required for M to move from A to its given position, then,
d=(at, (2)
and x = r cos 6 = r cos w^ (3)
Hence the velocity of P, the projection of M, is,
doc
v = --j:= — 03r sin o)t=—(j}y, (4)
and a = -^=—co^r cos cat =—a}^x (5)
Therefore, the motion of P is harmonic and the constant k in
equation (1) is here equal to co^.
This method of generating a simple harmonic motion is a con-
venient one for studying certain features of the motion. From
the definition it follows that a simple harmonic motion is a periodic
motion. In the study of harmonic motion certain terms are used
which are defined as follows: the amplitude is one-half the length
of the path of the point, that is, in this example, the radius of the
circle; the frequency is the number of complete oscillations of the
point per unit of time, equal, in this exa,mple, to the number of
revolutions per unit of time of the point M; the period is the
time required for one complete oscillation, that is, the time
required for M to make one revolution. Hence the period T is.
and the frequency ?i is.
'■=? <«>
L=ie_ (7)
ILLUSTRATIVE PROBLEM
305. A point moves with a simple harmonic motion such that its speed
is 90 in./sec. when it is 4 in. from the center of its path and 80 in. /sec. when
it is 6 in. from the center. Determine the period and the amphtude of the
motion. Determine also the maximum velocity and the maximum acceleration
of the point.
SIMPLE HARMONIC MOTION 255
Solution. — From equation (4) of Art. 126,
t; = co2/ = coVr2— a;2, (1)
Hence,
90 = coVr2_l6, (2)
and,
80 = coVr2-36 (3)
Dividing (2) by (3),
9_VV2-16
8~Vr2-36*
Squaring and transposing,
81(r2-36)=64(r2-16).
Hence,*
17r2 = 1892,
That is,
r2 = 111.3.
Therefore,
r = 10.56in.
Substituting this value of r in (2), we have,
to = 9.22 rad./sec.
From equation (6) of Art. 126,
Hence,
r=— .
CO
T=^ = .68sec.
9.22
The velocity at any point may be found from equation (1) . Since the velocity
is a maximum when a; =0, the maximum value is,
V = (ar
= 9.22X10.56 = 97.5 in./sec.
The acceleration is a maximum when the displacement of the point is greatest.
Hence, by using Equation (5) of Art. 126, the maximum value of the accelera-
tion is found to be,
= (9.22)2X10.56 = 897 in./sec.2
306. The drivers of a Mikado locomotive are 60 in. in diameter and the
length of the crank is 15 in. If the speed of the locomotive is 30 mi. per hr.,
determine the maximum velocity and the maximum acceleration of the cross-
head and piston relative to the engine frame, assuming that the connecting
rod is so long that the motion of the crosshead is harmonic.
Ans. V = 22 it. /sec; a = 387 ft./sec.^
256 MOTION OF A PARTICLE
307. A point moves with a simple harmonic motion the amplitude of
which is 10 in. If the period is 2 sec, determine the maximum velocity and
maximum acceleration.
308. The maximum velocity of a point which has a simple harmonic motion
is 10 ft. per sec. and the period is I sec. Determine the amplitude of the
motion and the maximum acceleration.
127. Non-uniformly Accelerated Motion. — One special case
of non-unifornily accelerated motion is treated in the preceding
article. Each case of this type of motion presents details peculiar
to itself, but the fundamental equations expressing the relations
between the displacement, velocity, acceleration, and time*are the
same for all cases. For rectilinear motion the equations are,
ds ...
. '=dt' • ■ (1^
dv d^s ' .-,
"^irw^ (2)
Eliminating t from these two equations, we have
ads = vdv (3)
By integrating equation (1), the following equations are obtained:
I ds= \ vd(j
J SI . Jll
or.
S2—si = As= I vdt, (4)
and,
\''dt= (''his,
Jn Js, V
or,
rs^ 1
t2-U = M= -ds (5)
By integrating equation (2), the following equations are obtained:
XP2 rt2
dv= I adt,
or,
1^2 — Vi = Ai;= J adi (6)
Jii
NON-UNIFORMLY ACCELERATED MOTION 257
and,
J I 'dt= I -dv,
or,
i2-h = At= i'^-dv (7)
Jvx a
The above equations may also be used in the case of curvilinear
motion if the tangential component of the acceleration is used, since
dv
(it=-T.' The equations may also be used if only a component, in
dz
a given direction, of the motion is considered, provided that s, v, and
a in the above equations be replaced by the components of displace-
ment, velocity, and acceleration, respectively, in the given direc-
tion. Thus,
_dvx_d^x
""'"dt'W'
a^/dx = Vxdvx, etc.
The relations between the displacement, velocity, acceleration,
and the time, may, in general, be determined from the above equa-
tions by either of two methods; namely, by solving the equations
by the methods of calculus, or by a graphical method through the
use of distance-time, velocity-time, and acceleration-time graphs.
The use of distance-time and speed-time graphs in finding the
speed and acceleration, respectively, have been discussed in
Arts. 113 and 119. The more convenient of the two methods
for any given problem depends upon the character of the problem.
Calculus Method. — Equations (1) to (7) may be solved by the
methods of calculus provided that certain relations between
the variables are known. Thus, if s is expressed as a fimction of i,
the velocity v may be obtained from equation (1). Likewise,
if V is expressed as a function of s, the time required for a given
displacement may be obtained from equation (5), and so on.
ILLUSTRATIVE PROBLEMS
309. A point moves along a straight path according to the law, v=3t^+4,
the iinits of distance and time being the foot and second, respectively. If
s =0 when t = 0, what is the value of s when t = 10 sec?
258 MOTION OF A PARTICLE
Solution. —
Hence,
ds P*
v = — or s= I vdt.
dt Jn
rio r "1 10
(= j (Zt^+4:)di = h*+4n = 1040 ft.
310. A point moves along a curved path according to the law ar = 10<+5.
If y = 10 ft. /sec. when t = 4 sec, what is the value of v when t = 7 sec?
Solution. First Method:
at = j = 10t+5,
Hence,
Av= I aidt= I {l0t-\-5)dt
'= jaidt= M
= r5/2+5<l'=180ft./sec
Hence, the gain in speed in the interval between the end of the fourth second
and the end of the seventh second is 180 ft. /sec and, therefore, the speed at the
end of the seventh second is,
D7 = 10+180 = 190ft./sec
Second Method. — Instead of using a definite integral as above, the problem
may be solved by means of an indefinite integral as follows :
=/a^=J(
v= jatdt= \ {I0t+5)dt
The constant of integration may be determined by means of the initial condi-
tion that y = 10 ft. /sec when f = 4 sec. Hence, using this condition, the above
equation becomes,
10 = 5X42+5X4+C.
Therefore, C= — 90 and the velocity at any time may be obtained from the
equation,
v = 5t^+5t-90.
If < =7 sec, the corresponding value of v is,
V7 = 5X7H-5X7-90 = 190 ft./sec
811. A -jpomt starts from rest at the vertex of the parabola y^ = x and moves
along the parabola according to the law x = —,x and y being in feet and t in
seconds. Find the magnitude of the velocity and of the acceleration of the
point at the end of 2 sec
NON-UNIFORMLY ACCELERATED MOTION
259
Solution.— The axial components of the velocity and of the acceleration
may be found from the equations of Art. 122. Thus,
t;«=-3T = -jrl-jl =2 = 1 ft./sec. when t = 2sec,
dy d(t\ 1.^ ,
dvx d (l\ 1 fi. / o
dVy d /1\ _
= V(l)2+(i')^ = 1.12 ft./sec.
= ^(i)'+(0)2 = 0.5 ft./sec.2
Therefore.
and,
312. The length of the crank, OA, of a steam engine (Fig. 297) is denoted
by r and the length of the connecting rod, BA, is denoted by I. If the crank
Fig. 297.
turns with constant angular velocity, to, determine the velocity and the accel-
eration of the crosshead, B, in terms of r, I, c<},*and the angle, ^, which the crank
makes with the horizontal.
Solution. — ^The displacement, s, of the crosshead from its extreme position
may be determined in terms of 9, and sinc0 6 = (ai\s may also be expressed as a
function of t, since w is known. Thus,
But,
and.
s = l-{-r — l cos <f>—r cos 6.
AC = l sin <f> = r sin 0,
I cos (t> = Vl^-l^sm^ 4> = Vi2_^2 sin2 e.
260
MOTION OF A PARTICLE
By expanding the last expression and using only the first two terms of the
T
expansion, since y is generally small, the last equation may be written, with a
close degree of approximation,
Therefore,
Hence,
I cos <t> = l — :^j sin2
r—rcosd+^j sin2 d.
ds
dd ,r^ . ^
-jl-r-r sin 0 cos
dt I
dd
dt
and,
(■
rw (sm d-\—, sm d cos 6
)■
(
cos e-\-rj (cos2 0— sin2
«)]^
rco^ j cos ^+y COS 26
)-
Graphical Method. — The solution of equations (4) and (6) by
the aid of speed-time and acceleration-time graphs will now be
shown. The speed-time graph has already been defined (Art.
119). The acceleration-time graph is a curve, the coordinates
of any point of which represent simultaneous values of the accel-
eration and the time. The acceleration used is either the total
acceleration if the point has rectilinear motion, or the tangential
acceleration if the point moves on a curved path.
Time, sec.
Fig. 298.
In Fig. 298 is represented a speed-time curve for a certain
rectilinear motion. The area under the curve between the ordi-
nates vi and V2 is expressed by.
Area
= I vdt.
NON-UNIFORMLY ACCELERATED MOTION 261
But by equation (4), As= I vdt.
Therefore, the area under the speed-time curve represents, to some
scale (depending on the scales used in plotting the graph), the dis-
tance traveled during the time interval t2 — ti, corresponding to a
change in the speed from vi to t;2.
ILLUSTRATIVE PROBLEM
313. The speed-time curve for the rectilinear motion of a certain point is
shown in Fig, 298. The scales are: 1 in., vertically, equals 200 ft. /sec, and
1 in., horizontally, equals 20 sec. If the area under the curve between i = 10
and i = 35 is 1.2 sq. in., how far does the point travel in the interval?
Solution. — 1 sq. in. represents 200 ft. /sec. X20 sec. =4000 ft.
Hence, the distance =1.2X4000 = 4800 ft.
Let the curve in Fig. 299 represent the acceleration-time graph
for the rectiUnear motion
of a point. The area
under the curve between ^ soo
the ordinates ai and a2 is «
expressed by, «2oo
adt. e
But, by equation (6),
Av
= I adt.
Jti
100- -
Fig. 299.
Therefore, the area un-
der the acceleration-time curve represents, to some scale (depend-
ing on the scales used in plotting the graph), the change in speed
during the time interval t2 — ti.
ILLUSTRATIVE PROBLEM
314. The scales used for the acceleration-time curve in Fig. 299 are:
1 in. = 200 ft, /sec. 2 and 1 in. = 2 sec. If the area under the curve between the
ordinates t = 1.5 sec. and t = 4: sec. is 1.3 sq. in., what is the speed at the end of
4 sec. if the speed at the end of 1.5 sec. is 20 ft. /sec?
Solution. — 1 sq. in. represents 200 ft./sec.2X2 sec. =400 ft. /sec.
262
MOTION OF A PARTICLE
Hence,
Therefore,
Ay = 1.3X400 = 520 ft./sec.
^4 =20+520 = 540 ft./sec.
PROBLEMS
315. A point moves on a circular path having a radius of 10 ft., according
to the law, v = 20t. Determine the position of the moving point, with respect
to its initial position, at the
end of 5 sec.
316. A bellcrank oper-
ated by a cam (see Fig.
280) rotates so that its
angular acceleration a fol-
lows the law, a = 4^2 +6. If
its initial angular velocity is
2 radians per sec, what is
the angular velocity of the
bellcrank at the end of
0.5 sec?
Ans. (0 = 5.17 rad.ysec.
317. Two railway sta-
tions are connected by two
straight parallel tracks.
Two trains A and B start
from rest at one station
and reach the second sta-
tion in 150 seconds. The
speed-time curve for A
is shown in Fig. 300(a)
and that for B in Fig. 300(6) . Draw the acceleration-time curve for each train,
318. What, in Fig. 300, represents the distance between the stations?
From the diagrams find the distance in miles. Ans. 1.5 miles.
128. Relative Motion. — In the preceding articles, the motion
of a particle is defined or described with reference to a point or a set
of axes assumed to be fixed. All points and bodies, however, are
in motion, but in most practical problems it is convenient to con-
sider the earth to be fixed. Therefore a set of axes passing through
any point on the earth will be regarded as a fixed reference frame.
The motion of a particle, described with reference to a point on the
earth, is called its absolute motion. The motion of a particle,
described with reference to a point that is moving with respect
to the earth, is called its relative motion. It will be observed that
80 100
Time, sec.
Fig. 300.
RELATIVE MOTION
263
the absolute motion of a particle is its relative motion with respect
to the earth.
In considering the motion of two moving particles, the motion
of one particle relative to the other particle is often required.
The relation between the absolute and relative motions of two
particles A and B (assumed, for convenience, to be moving in a
plane) may be stated by the following important theorem.
The absolute displacement, velocity, or acceleration of A is the
geometric or vector sum of the relative displacement, velocity, or accel-
eration, respectively, of A with respect to B and the absolute dis-
placement, velocity, or acceleration, respectively, of B.
This theorem may be expressed in the form of equations as
follows:
SA = SA-B>Sfl or Sa = Sa-^Sb*
B B
Va = Va^Vb or Va = Va-^Vb.
B B
a^ = cIa -B' ttB or a^ = aA—^B»
Y
B ^B °i
A,
/sa
1 ¥
Fig. 301.
Fig. 302.
where Sa denotes the absolute displacement oi A, Sb denotes the
absolute displacement of B, and s^ denotes the relative displace-
B
ment of A with respect to B, and similiarly for velocities and
accelerations.
Relative Displacement. — As applying to displacements the above
theorem is nearly self-evident. To illustrate, let A and B be two
particles which occupy the same position with reference to a set of
264
MOTION OF A PARTICLE
axes as shown in Fig. 301. Let tlie point A be given a displace-
ment Sa = AAi and the point B, a displacement Sb = BBi. The
point A moves relative to the point B a distance BiAi and in the
direction BiAi. Hence BiAi is the displacement of A relative
to B and it is evident that,
Sa = Sa-HSb or Sa = Sa'^Sb .
If the points A and B do not occupy the same positions, the
reasoning is changed but Uttle, the final conclusion being the same
as stated above as will be evident from a study of the diagram in
Fig. 302.
ILLUSTRATIVE PROBLEM
319. The current in a river with parallel sides is 4 mi./hr. A motor boat
starting from one side keeps headed perpendicular to the sides and moves at
6 mi./hr. If the river is one mile wide, what is the absolute displacement of
the boat after reaching the other side?
Solution— Let B (Fig. 303) rep-
resent the boat and W the water
that is in contact with the boat when
the boat starts. Then,
S^ = Sw +> Sjf .
w
^FiG. 303.
The displacement of the water while
the boat is moving across is,
s^ = WWi = l mile.
After the boat has reached the other side its displacement relative to the
water (now at Wi) is,
Sg =1^151 = 1 mile.
w
Therefore,
S5 = H^ ! = V(l)2+(|)2 = 1.2 miles,
and,
tan e
w_3
(9 = 56° 20'.
Relative Velocity. — The above theorem, as applying to the
velocities of two moving particles, may be proved as follows:
RELATIVE MOTION
265
Let A and B (Fig. 304) be any two moving particles (assumed in a
plane), let the absolute velocity, Va, of A be represented by AAi,
and let BBi represent the abso-
lute velocity, vb, of B. Let a
velocity equal to Vb but reversed
in direction be given to each
point. The point B will then be
at rest and, hence, the resulting
velocity of the point A is the
velocity of A relative to B. Hence,
it is evident from the diagram that,
■^'b
Va = Va-¥^Vb-
Fig. 304.
In this proof it is assumed that the additions of equal velocities to
each particle does not influence the relative velocity of the two
particles. This assumption is in accord with an important prin-
ciple in kinematics, namely, that the relative motion of two par-
ticles is not affected by any motion that they have in common.
The above equation as already noted, may be written in the
form
Fig. 306.
VA-o}r =
which states that the relative velocity
of two points is the difference of
their absolute velocities, as is shown
in the diagram.
ILLUSTRATIVE PROBLEM
320. In the shaper mechanism shown in
Fig. 305, let A be the shding block and let
B be the point on the rocker arm OSM
which is coincident with A at the instant.
The distance OOi is 20 in. and d is 30°. If
CO = 20 r.p.m. and r = ^ ft., find vb, the abso-
lute velocity of B, and Va, the relative
velocity of A with respect to B.
Solution. — Since the block A moves on
a circular arc at constant speed, its velocity
is given in magnitude by the equation,
20X27r,,l .f.-f.,
— go — X2 = J05 ft./sec.
266 MOTION OF A PARTICLE
and its direction is perpendicular to r, as shown in Fig. 305. The direction of
the absolute velocity of B is perpendicular to OiBM and its magnitude is
unknown. Likewise, the direction of the velocity of A relative to B is known
since it is along (parallel to) the rocker arm. By applying the equation,
B
the magnitudes of Va and vb are determined by the intersection of the lines
"b
ir directions. By scaling off the magnitu
VA=0.70ft./sec. and V5 = 0.79 ft./sec.
B
that represent their directions. By scaling off the magnitudes, the following
values are found,
PROBLEMS
321. Two trains A and B travel on parallel straight tracks. The speed of
A is 40 mi./hr. and that of B is 50 mi./hr. in the same direction. What is the
velocity of A relative to B1 Of B relative to A?
Fig. 307.
322. A train A travels with a velocity va =40 mi./hr. and another train B
travels with a velocity vb = 50 mi./hr. in the directions shown in Fig. 306.
What is the magnitude and the direction of the relative velocity of B with
respect to ^? Ans. 55.4 mi./hr.
323. In Fig. 307, let A be a block which revolves about 0 at a constant
angular velocity co = 30 r.p.m. and let B be the point on the arm OiBM coin-
cident with A at the instant. If r = OA --= 9 in ., OOi = 3 in ., and 6 = 4:5°, find the
absolute velocity of B.
Relative Acceleration. — In order to prove the above theorem as it
relates to accelerations, it will be assumed that the relative accel-
RELATIVE MOTION
267
eration of two particles is not changed if equal additional accel-
erations be given the two particles. This is analogous to the
assumption made in the discussion of relative velocities. Hence, if
equal velocities and accelerations be given to each of two particles
the relative motion of one with respect to the other will not be
changed. Consider, then, two particles A and B which move along
paths as indicated in Fig.
308. Assume that a veloci-
ty equal and opposite to the
velocity of B{ — Vb) is im-
posed on each particle (not
shown at A in Fig. 308).
Assume also that acceler-
ations equal to the acceler-
ation of B but of opposite
sense are imposed on the
two particles. This will
not change the relative
motion of A with respect
to B. Imposing a velocity
of —Vb and an acceleration
of —Gb on the particle B will make its position fixed, and hence
the acceleration of A with respect to B will be the same as the
absolute acceleration of A, The acceleration of A, however, is
now the vector sum of a^ and —Qb and hence.
Fig. 308
or,
Ca -^ Qb-
Hence, the relative acceleration of one point with respect to a
second point is the vector difference of the absolute accelerations
of the two points.
ILLUSTRATIVE PROBLEM
324. Two friction disks (Fig. 309) rotate with constant angular velocities.
Let A and B be points on the circumferences of the large and small disks,
respectively. If coi = 40 r.p.m., ri=9 in., and r2 = 3 in., find the magnitude
and the direction of the acceleration of A relative to B.
8
MOTION OF .
SolviiAm.—'
Wiri = (aiTi:,
coiri 40X9
120 r.p.m.,
aA = coi V, = (^^) ' X ^ = 13 . 3 ft./sec.2,
/120X27r\2 3
[—m-J Xj2 = 40ft./sec.^.
Ob = iJi'-^r-i ■■
Therefore,
and,
Therefore,
(lA =aA—^aB
B
= V(i3.3)2+(40)2 = 42.2 ft./sec.2,
tan<^ = ^ = i5_§=0.33.
as 40
0 = 18° 25'.
PROBLEMS
325. What is the relative acceleration of Ai (Fig. 309) with respect to B,
the data being the same as in the
preceding problem?
326. Two trains A and B
travel in the same direction on
parallel tracks; A increases its
speed uniformly 10 mi./hr./min.
. o and B decreases its speed uni-
J ^j[ ^ formly 5 mi./hr./min. What is
the relative acceleration of A with
respect to B?
Fig. 310.
COMPONENTS O'F ACCELERATION
269
327. An automobile, A, is traveling on a straight road, increasing its speed
at the rate of 300 ft./min./sec. when in the position shown (Fig. 310). At
the same time another automobile, B, is traveling in a circular path increasing
its speed at the rate of 5 ft. /sec. ^ Its speed when in the position shown is
vb = 12 mi./hr. and the radius of the circular path is 40 ft. What is the rela-
tive acceleration of A with respect to 5 ? Ans. 12.3 ft. /sec. ^
129. Transverse and Radial Components of Acceleration. — In
addition to the x- and ^/-components and the n- and ^-components
of the acceleration of a particle (Art. 122), the transverse and
radial (T and R) components are sometimes convenient to use.
The transverse direction is perpendicular to the radius vector
Fig. 311
drawn from any point taken as the center or pole, and the radial
direction is along, or parallel to, the radius vector (see Fig. 311).
The T- and 7^-components may be derived by the application of
Theorems I and II (Art. 118). In Fig. 311(a) let m represent a
particle moving on a fixed path. Let 0 be any center or pole
(not the instantaneous center of rotation, Art. 135). Then, the
total velocity, v, of the particle may be replaced by the tM'^o com-
ponents Vt and Vr. Thus,
Fig. 311(a) indicates that in general at any instant
Vt is changing in magnitude and also in direction, and that,
Vr is changing in magnitude and also in direction.
Hence there will be four components of the acceleration: two of
270 MOTION OF A PARTICLE
the type -t:; and two of the type voo, as follows (shown in Fig.
3116):
change of the magnitude of Vr.
(2) Vrco perpendicular to vr, being the rate of change of Vr
due to the change in its direction only.
the rate of change of the magnitude of Vt.
(4) Vt03 = co^p = — - perpendicular to Vt, being the rate of
P
change of Vt due to a change in its direction only.
The T- and /^-components of the total acceleration are,
then (Fig. 26),
aT = VR03-{-{vRCx)-\- pa) = 2vR0)-\-pa = 2co --r,-\- P~Tf^~ 'tXp^^)}
dvR d?p 2
130. Coriolis* Law. — It is convenient to consider the motion
of a point of a body as being generated by a motion of the point
along a path (line) as the path moves. Thus, in plane motion
of a rigid body the motion of any point in the body may be con-
sidered as a combination of a motion along a path and a motion
due to a translation of the path.
If a point moves along a path as the path is translated, the
acceleration of the point is the vector sum of the acceleration
relative to the path and the acceleration of the point of the path
which is coincident with the point at the instant (or of any point
of the path since the path is translated). This leads to the same
expression for the acceleration of the point as was obtained in Art.
128, as wiU be shown later.
If, however, a point moves along a path as the path rotates,
the acceleration of the point is obtained as the vector sum of
three component accelerations. The relation between these three
component accelerations and the total acceleration is known as
Coriolis' law, which may be derived by use of the theorems of
Art. 118 as follows:
CORIOLIS' LAW
271
In Fig. 312(a), let m represent a particle (or bead) traveling
along a path (curved wire) with the velocity u, and at the same
time let the path rotate about 0 with angular velocity co. The
velocity of the particle will then have, at any instant, two com-
ponents, u and w, u being the velocity along (relative to) the path,
and w the velocity of the point on the path which, at the instant,
coincides with the particle.
It will be noted from Fig. 312(a) that, as the particle passes
from one position to some other position, the velocity u changes in
magnitude and also in direction. Likewise w changes in magni-
Fig. 312.
tude and also in direction. At any instant, therefore, the accel-
eration of the particle 7n will have at least four components. It will
be found convenient, however, to consider the change in both u
and w to be made up of several parts, as follows:
The change in u from ui to U2 is made up of:
(1) A change in magnitude, due to an increasing (or
decreasing) speed along the path.
(2) A change in direction, due to the curvature of the path.
(3) A change in direction, due to the turning of the path.
The change in w from w\ to W2 is made up of:
(4) A change in magnitude, due to the rotation of the path
with increasing (or decreasing) angular velocity
272 MOTION OF A PARTICLE
(assuming p to remain constant, tliat is, considering
the particle to be fixed to tiie path).
(5) A change in direction, due to the rotation of the path
(considering the particle fixed to the path) .
(6) A change in magnitude, due to the change in length of p
(assuming co to remaia constant), caused by the
movement along the path.
(7) A change in direction, due to the change in the direc-
tion of p, caused by the motion of the particle along
the path (assuming the path fixed). The change in
direction of p and of w are the same since they remain at
right angles.
At any instant, therefore, the acceleration of m will have seven
components, unless (4) and (6) are combined, three being of the
type -j- and four of the type vo3. These components (repre-
ill
sented in Fig. 3126) are expressed as follows:
nlJj
(1) --j- — ra parallel to u and having the same sense as ii
since fi, the angular velocity, and a, the angular accel-
eration, of the particle m with respect to the center of
curvature, S, of the path were assumed to agree in sense
as indicated in Fig. 312(a).
(2) uQ, = V:^r = — perpendicular to it, towards S.
(3) uo) perpendicular to u toward S.
(4) -tt = per parallel to w and having the same sense as w,
di
since co, the angular velocity, and cr, the angular
acceleration, of the path with respect to 0 are assumed
to agree in sense.
(5) wo3 = o?p = — perpendicular to w, toward 0.
By referring again to Fig. 312(a), it will be noted that the velocity
u could also be produced by allowing the particle (bead) to slide
along the straight Hne p as the line p is rotated about the center 0
with the same angular velocity, co, as that of the path. The two
components of u would then be Vt= cop and Vp=-Tr (see Fig. 312c).
CORIOLIS' LAW
273
It will be noted that these components are the radial and trans-
verse components discussed in Art. 116. We are concerned at
present only with the changes in Vt due to this motion of the par-
ticle along the path, since the changes in w, not already con-
sidered in (4) and (5), are due to the motion of the point along the
path and are the same as the changes in Vt. Now Vt changes in
magnitude and also in direction; these changes correspond to the
change in length and in direction of p. The accelerations thus
produced by reason of the change in vt (or w) are then;
dvr d{o)p) dp
(6)5f =
03 --7-=VpOi, parallel to Vt (or w) and oppo-
dt dt ^ dt
site to the sense of w since w is decreasing. And,
(7) Vroi perpendicular to Vt (oi w) toward 0.
In (4) p was considered constant and w was assumed to change
whereas in (6) co was considered constant while p was assumed to
change. Of course (4) and (6) are parts of one expression, namely,
if — P ~Tf~^^ 'Tfy where co and p are considered to change at
the same time. Separating the operation into two parts, how-
ever, makes clear the need for both co and p to vary.
Finally, the total acceleration is expressed in terms of the seven
components as follows :
a = ra-f> {
1/2
r
-Ht/«-H'p(r+)>
Wta
up
T
-H^ mVp -H (aVt,
These components may be grouped in several ways, the most use-
ful grouping depending upon the particular problem under consid-
eration. One such grouping leads to Coriolis' law. By combining
the last two components into one term, we have oi{Vp->r^VT)=uu3
since Vp-\^VT = u (Fig. 312c). And since uoi represents an accel-
eration due to a change in the direction of u it must be perpen-
dicular to u (Theorem II, Art. 118). This may be added to the
third component, since the two will always have the same sense.
Hence the third, sixth, and seventh components combine into
2wco directed perpendicular to u with a sense such that if it were
applied as a force at the end of the vector u it would cause the
Vector u to turn in its actual direction of rotation. Further, let ar
274
MOTION OF A PARTICLE
represent the vector sum of the first two components and let am
represent the vector sum of the fourth and fifth, that is, let
u^ , If-
ar=ra -^ — and a„,= per -f> — .
r p
The total acceleration, then, may be expressed as follows.
This equation maybe stated in words as follows: If a particle moves
on a path as the path rotates, the acceleration of the particle is the
geometric or vector sum of (1) the acceleration the particle would
have if the path were fixed, (2) the acceleration the particle would
have if it were fixed on the path as the path rotated, and (3)
2m CO called the compound supplementary acceleration. This state-
ment is known as Coriolis'' law.
From the above derivation, it is seen that 2uw is due (1) to a
part of the change of direction of u, (2) a part of the change of the
direction of w, and (3) a part of the change of the magnitude of w,
all of which would be absent if the path were translating instead of
rotating. Thus, as noted at the beginning of this article, if the path
is translated, the equation a = a^ -B- a,„ expresses the same fact as
does the equation of Art. 128 which involves accelerations,
ILLUSTRATIVE PROBLEM
328. The circular arc APB (Fig. 313a) represents the vane of a centrifugal
pump, P being a particle of water. Find the acceleration of the particle P
Scale 1 1^200 ft. per sec?
Fig. 313.
when it is 12 in. from 0, the center of the shaft, if, at that instant, the angular
velocity of the wheel is 10 rad. per sec. clockwise and its angular acceleration
is 50 rad. per sec.^ clockwise. The tangential velocity of the particle along
(relative to) the vane is 10 ft. per sec. and the tangential acceleration relative
CORIOLIS' LAW
275
to the vane is 10 ft. per sec.^ OB makes an angle of 45° with the horizontal
and is 18 in. long; OA is 3 in.; and OP is 12 in. The radius of the arc APB
is 13J in.
Solution. — By Coriolis' law the total acceleration a is,
a = Or -i^ dm ~^ 2uco,
ar and Om are found most easily from their t- and n-components, Hence,
a = (aT)i 4> (ar)„ 4> {am)^ -f> («»,)„ +>• 2Maj
102 12 12
= 10 4>:f^4>50X^+>102X^+> 2X10X10
I04 xZ \.Z
I2"
= 10 -+> 90.56 -f> 50 +> 100 4> 200,
each of the quantities being expressed in ft. per sec. 2 The five components
are shown in their proper directions in Fig. 313(a) and the resultant accelera-
tion a as found from the acceleration polygon is shown in Fig. 313(6). By
scaling the closing line of the polygon, a is found to be 145 ft. per sec.^ in the
direction shown in Fig. 313(6).
PROBLEM
329. Point P (Fig. 314) moves on the rod OM which rotates about the
fixed point 0. In the given position OP = 6 ft.; the
velocity along the rod is 5 ft. per sec. toward M and is
increasing at the rate of 5 ft. per sec. each second. The
angular velocity of the rod is 2 rad. per sec. in a clockwise
direction, and is decreasing at the rate of 1.5 rad. per sec.
each sec. Find the acceleration of the po'nt P. Note. —
Usescales: lin. = 2ft.; 1 in. = 5 ft. per sec; 1 in. = 5 ft. per
sec. 2 Use 8| in. by 11 in. paper with the long edge as the
bottom. Lay the diagram out on left part of sheet, placing
point 0 1 in. from the left edge and 2 in. from the bottom
edge. Show all component accelerations of P to scale.
Combine the components graphically, starting at a point 1 j in. from the top
edge and 1 in. from the right edge. Combine the components in the order
of (am)«> (Op dr and 2uco.
Ans. ap = 22.0 ft. /sec. ^ vertically downwards.
Fig. 314.
CHAPTER VIII
MOTION OF RIGID BODIES
131. Introduction. — In the preceding chapter the motion of a
point or particle and the relation between the motions of two
particles have been considered. In engineering problems in
general, however, the motion of bodies, not particles, must be
considered. In some problems, the dimensions of the body may
be assumed to be negligible in comparison with its range of motion,
with very small error, and hence, the methods and equations of
the preceding chapter apply directly to the motion of the body.
This assumption is involved in a number of problems in the pre-
ceding chapter, in which a body is assumed to be a particle. In
general, however, the motion of bodies as met in engineering prac-
tice is such that the various points of a body have different
motions.
The object of this chapter is to analyze certain common types
of motion of rigid bodies so that the displacement, velocity, and
acceleration, both linear and angular, may be found from the
methods and equations developed in the preceding chapter. The
motions considered are translation, rotation, and plane motion.
132. Translation. — Translation of a rigid body is a motion
such that no straight line in the body changes direction, that is,
each straight line remains parallel to its initial direction. Hence,
at any instant, all points in the body move along parallel paths
and have the same velocity and acceleration. If the transla-
tion is such as to move the particles on curved paths the motion is
called curvilinear translation as, for example, the motion of the
parallel rod of a locomotive. If the particles move on straight-
line paths the motion is called rectilinear translation. The body
may have a uniformly or non-uniformly accelerated motion.
Hence any point in a translating body may have any of the motions
treated in the preceding chapter. Further, since all particles in
the body have the same motion at any instant, the displacement,
velocity, and acceleration of the body is described by the dis-
placement, velocity, and acceleration of any particle of the body.
276
ROTATION 277
PROBLEM
330. A locomotive is running on a straight track at a constant speed of
40 mi./hr. The diameter of the drivers is 6 ft. and the radius of the crank-
pin circle is 15 in. What is the magnitude and the direction of the velocity
Fig. 315.
and of the acceleration of the parallel rod, relative to the engine frame, when
the rod is in the position shown in Fig. 315? What is the absolute velocity
and the absolute acceleration for the same position?
133. Rotation. — Rotation of a rigid body is a motion such that
one line in the body (or body extended) remains fixed in space
while all points of the body describe circular paths having centers
on the fixed line. The fixed line is called the axis of rotation and
the plane in which the mass-center of the body moves is called
the plane of motion. The point of intersection of the axis of
rotation and the plane of motion is called the center of rotation.
It will be noted that any line parallel to the plane of motion
changes direction. The motion of a body having rotation cannot
be defined or described by stating the linear displacement, velocity,
and acceleration of any point in the body, as was the case for trans-
lation, since all points in the body do not have the same linear mo-
tion. However, the angular displacements, velocities, and acceler-
ation, respectively, are the same for all particles in the body. Hence
the motion of a rotating rigid body may be described by the
angular motion of any point in the body. Thus all the equations
in the preceding chapter deaUng with the angular motion of a
point moving on a circular path, in which the radius vector is the
radius of the circle, apply to the motion of a rotating rigid body,
as well as to each point in the body. The linear displacement,
velocity, and acceleration of any point may also be found from the
equations in the preceding chapter, that deal with the linear
motion of a point moving on a circular path.
278 MOTION OF RIGID BODIES
PROBLEMS
331. A straight stick 4 ft. long rotates in a horizontal plane about a ver-
tical axis through one end of the stick. Its angular velocity changes uniformly
from 20 to 50 r.p.m. in 5 sec. What is the linear velocity of its mid-point at
the end of 2 sec?
332. The flywheel of a punching machine fluctuates from 100 to 80 r.p.m.
at a uniform rate when a hole is punched. If the flywheel makes 1| revolu-
tions while this change of speed takes place, how long does it take to punch
the hole? Ans. t = l sec.
333. The flywheel of a rolling-mill engine is 14 ft. in diameter. Just before
the steel is fed in the rolls the speed of the flywheel is 90 r.p.m. As the
steel enters the rolls the speed decreases uniformly during | sec, before the
governor can operate. If the angular acceleration (negative) of the flywheel
is 20 r.p.m. per sec, what is the decrease in the speed of the flywheel, expressed
in r.p.m.?
134. Plane Motion. — Plane motion of a rigid body is a motion
such that each point in the body remains at a constant distance
from a fixed plane. The motion of the connecting rod of a steam
engine is an example of plane motion. The wheels of a locomotive
when running on a straight track also have plane motion. A
plane parallel to the fixed plane, containing the mass-center of the
body, is called the plane of motion. It is evident that a rotation
is always a special case of plane motion, whereas a translation
may or maj^ not be a plane motion.
In plane motion, in general, a straight line in the body lying in
the plane of motion changes direction and, hence, the body rotates,
but not about a fixed axis. The body, therefore, has angular mo-
tion, and its angular displacement, velocity, and acceleration are
the same as that of any straight line in the body, in the plane of
motion, since aU such lines have the same angular motion if the
body is rigid. The angular motion of the body, therefore, may be
studied by means of the same equations that apply to the rotation
of a rigid body about a fixed axis.
Rotation, however, is only one part of the motion of a rigid
body having plane motion (except in a special case as noted in the
next article). Plane motion of a rigid body may be resolved into
two component motions, a rotation and a translation, according
to the following theorem:
Plane motion of a rigid body, at any instant, is a com-
bination of: (1) a pure rotation of the body, about an axis
PLANE MOTION 279
(perpendicular to the plane of motion) passing through any
point B in the body, with an angular velocity and acceler-
ation the same as that which the body has at the instant;
and, (2) a translation of the body which gives to each point
the same linear velocity and acceleration that the point B
has at the instant.
The point B is called the base point. It is evident that all
points except the base point have two motions; a rotation about
the base point, and a motion the same as that of the base point.
From the analysis of the motion according to the above theorem,
the displacement, velocity, and acceleration of any point, A, in the
body may be found from the equations developed in Art. 128 of
the preceding chapter. The equations are,
B
Va = V^^Vb
B
c^A = «^ -f> as.
To arrive at the above theorem, consider the motion of the
connecting rod of a steam engine (Fig. 316). Let P denote the
position of the crosshead and Q that of the crank-pin, the crank
being of length OQ, When the crank moves from position OQ
Fig. 316.
to position OQi the connecting rod moves from position PQ to
position PiQi. This change of position can be given the rod by
first rotating the rod about P, until it becomes parallel to its new
position, and then giving the rod a translation such that each
point receives a displacement equal to PPi. By this combina-
tion of motions the point P moves along its actual path but any
other point does not travel in its actual path. The point Q, for
280 MOTION OF RIGID BODIES
example, moves along the path QQ'Qi instead of its actual circular
path QQi. However, as the change of position is made smaller
and smaller, the path QQ'Qi approaches the circular path QQi
and, in' the limit, as the two motions are generated simultaneously,
each point is made to move on its actual path by successive
combinations of a proper rotation and a proper translation. The
rotation, at any instant, must give the bpdy its actual angular
velocity and acceleration at the instant, since the translation
does not influence the angular motion of the body. The transla-
tion must give all points of the body the same motion that the
base point has at the instant, since the base point receives its total
motion from the translation.
By the above method, the rod is given a rectilinear transla-
tion, since the point P moves on a straight-Une path. If, how-
ever, another base point is chosen as, for example, the point Q,
Fig. 317.
the change of position from PQ to PiQi may then be made by:
(1) a rotation about Q (Fig. 317); and, (2) a curvilinear transla-
tion giving to each point the same motion that Q has. By this
combination of a rotation and a translation the point P is made to
take the path PP'Pi instead of its actual path PPi, but, reasoning
as above, if the change of position is made smaller and smaller
the path PP'Pi approaches the path PP\ and, in the hmit, as
the two motions are imposed simultaneously, each point in the
body is given its exact motion at any instant. Likewise, any other
point in the body may be selected as the base point. And, the
plane motion of any other rigid body may be treated in like
manner. Hence, a plane motion of a rigid body may be consid-
ered, at any instant, as a combination of a rotation of the body
about any base point in the body, with an angular velocity and
acceleration equal to the angular velocity and acceleration that
the body has at the instant, and a translation of the body that
PLANE MOTION 281
gives to each point the same linear velocity and acceleration that
the base point has at the instant.
In applying the equations,
Vp = Vp-{^ Vq and ap = ap-[^ aq,
Q Q
to the motion of any two points in a rigid body having plane
motion, the point Q represents the base point. However, as stated
above, the base point may be any point in the body. Further, the
angular velocity (and acceleration) of the body is the same with
reference to all base points in the body.
Since any two points in a rigid body remain a fixed distance
apart, there can be no relative velocity of one point toward the
other, that is, the relative velocity of either point with respect
to the other is perpendicular to the fine joining the two points.
This is not true, however, of the relative acceleration, since it is
made up of a tangential and a normal component, the normal com-
ponent having a direction along the line joining the two points.
'The principles discussed in this article are involved in the solution
of the following problem.
ILLUSTRATIVE PROBLEM
334. A 50 h.p. engine has a cylinder 10 in. in diameter and a stroke of 10
in. The engine runs at a constant speed of co = 300 r.p.m. The ratio of the
length of the connecting rod to that of the crank is 5. Find the velocity
and the acceleration of the crosshead when the crank angle is 30°.
Fig. 318.
Solution. — Let P be the crosshead and Q the crank pin. Then in Fig.
318 PQ represents 25 in. and OQ or r represents 5 in., according to the scale
used. The velocity of P may be found from the equation,
Vp = Vp-\-^VQ.
Q
282 MOTION OF RIGID BODIES
There are six elements involved in the equation — three magnitudes and three
directions, four of which must be found before the equation can be used
to determine the other two. The direction of the velocity of Q is per-
pendicular to OQ and its magnitude is,
VQ = c^r= [ — -^ — j X j2 = 13 . 1 ft./sec.
The direction of Vp is perpendicular to the line joining P and Q and its mag-
Q
nitude is unknown. The direction of vp is horizontal. Hence by laying off
vq to a convenient scale in the proper direction from P, and by drawing a line
from the end of vq, perpendicular to the connecting rod, until it intersects a
horizontal Une through P, the magnitudes of vp and vp are determined by the
Q
intersection. By scaling off the values of vp and vp, the following results are
Q
found
«P = 11. 4 ft./sec. and vp = 7. 8 ft./sec.
Q
Thinking of the plane motion of the connecting rod as a combination of a
rotation and a translation, vp is the velocity which P is given by the rotation
Q
of the rod about Q and vq is the velocity given to P by the translation of the
rod. The two velocities produce the resultant velocity vp.
The acceleration of P is given by the equation,
ap=ap -\^ aq.
Q
For convenience, ap will be replaced by its tangential and normal components.
Q
Hence,
ap = /ap\ -f> (ap\ +>• ag.
XQ/t \ Q/ n
Eight elements are involved in the equation — ^four magnitudes and four
directions, six of which must be found before the graphical construction
representing the equation can be completed, uq is directed from Q toward
0 and its magnitude is,
„ /300X27r\2 5
OQ = co2r=(^ — ^^—j Xj2 = 412ft./sec.2
ap is known in direction, being parallel to PO (horizontal), since vp
changes in magnitude only. /ap\ is directed from P toward Q and its
\ qI n
magnitude is.
lvp\^
PLANE MOTION
283
{%):
has a direction perpendicular to PQ. The two unknown elements
are, therefore, the magnitudes of ap and / ap\ .
\ qI t
In Fig. 319, starting at P, the vectors are laid off to a convenient scale.
Thus aq and [ap\ (both of which are completely known) are drawn and
KqIu
then, fron the end of /ap\ , a line is drawn that represents the direction of
\ q) n
Fig. 319.
(ap\ , that is, it is drawn perpendicular to PQ. The intersection of this
qI t
line with the line PO (which represents the direction of ap) determines the
lengths of the vectors that represent iaA and a p. By scaling off the
\ q) I
values, the following results are obtained.
/apy = 200 ft./sec.
2 and ap = 404 ft. /sec. 2.
PROBLEMS
335. A wheel 4 ft. in diameter rolls, without slipping, on a horizontal track
(Fig. 320). The velocity of its center, at
a ^iven instant, is 4 ft./sec. to the right
and the acceleration of the center is 6
ft./sec .2 in the same, direction. Find the
velocities and the accelerations of the points
A and B.
336. Find, by the graphical method,
the velocity and the acceleration of the
point C of the four-link mechanism de-
scribed in Problem 337 (Fig. 322). Use
the following scales: lin. = lft.; lin.=2
ft./sec; and 1 in. = 10 ft./sec.^
Ans. ^0 = 1-54 ft./sec; ac =55.5 ft. /sec ^
284 MOTION OF RIGID BODIES
135. Instantaneous Center. — It was shown in the previous
article that a plane motion of a rigid body may be considered, at
any instant, as a rotation, about an axis through any point in the
body, combined with a translation. However, by choosing a par-
ticular axis in the body (or its extension), the motion of the body,
at any instant, becomes one of rotation only, that is, no transla-
tion need be combined with the rotation to produce the actual
motion of the body. This axis is called the instantaneous axis of
rotation or, the instantaneous axis of zero velocity. Its intersection
with the plane of motion is called the instantaneous center of
rotation.
To show a method of locating the instantaneous center, let A
and B be any two points in the plane of motion of a rigid body
having plane motion (Fig. 321). Let Va denote the velocity of A,
and Vb, the velocity of B. From A draw
a line perpendicular to Va and from B
draw a Hne perpendicular to Vb, the two
lines intersecting in 0. Any point in the
body on the hne OA, if not at rest,
must have a velocity perpendicular to
OA, since any two points in the body
Fig. 321. cannot move toward or away from each
other if the body is rigid. Likewise, any
point on the line BO, if not at rest, must move perpendicular to
BO. Hence, the point of intersection oi AO and BO must be
at rest since it cannot, at the same instant, move in two
directions. Therefore, the body, if not at rest, must rotate
about the point 0 at the instant.
If, then, A and 0 are two points in the rigid body having plane
motion, A being any point, and 0 the instantaneous center, the
equation,
Va = Va-^Vo,
o ,
reduces to,
Va = Va-^ 0=03X0 A,
d
in which w is the angular velocity of the body. Further, since co
is the same for all points in the body, it follows that the
velocities of any two points A and B vary as their distances
INSTANTANEOUS CENTER 285
from the instantaneous center. Or, expressed in equational
form,
VA^oiOA^OA
vb oi'OB OB'
It should be noted that the instantaneous center is the center of
zero velocity and not of zero acceleration. In the case of the
motion of rotation of a rigid body about a fixed axis, which is a
special case of plane motion, the axis of rotation is also the instan-
taneous axis and it has zero acceleration as well as zero velocity
since it is at rest. In the general case of plane motion, however,
the instantaneous center changes its position in the body and also in
space. Although the velocity of the point in the body, coinciding
with the instantaneous center at a given instant, is zero, the
velocity is changing through its zero value and hence the point
has an acceleration. Therefore in the equation,
a A = 0,^-^ do,
o
in which A is any point and 0 is the instantaneous center, ao is
not zero. Hence the actual acceleration, at any instant, of a
point in a rigid body having plane motion cannot be found by
considering the body to be rotating about a fioced axis through
the instantaneous center, as may be done in determining the
velocity of any point of the body.
ILLUSTRATIVE PROBLEM
337. The four-link mechanism, ABCD, shown in Fig. 322 has the following
dimensions: AB = Q in.; 5C = 3ft.; DC = 2ft.; AD = i ft.; 0 = 45°. Find the
instantaneous center for the link BC. If
the crank AB rotates at a constant angular
velocity w = 10 rad./sec, find the angular
velocities, co2, of the link DC and, cos, of
the link BC. Also find the linear veloc-
ity, vh, of H, the midpoint of BC.
Solution. — The instantaneous center
for the link BC is O, the point of inter-
section of the lines AB and DC extended.
By scaling off the lengths of OB, OC, and -p ^oo
OH, the following values are foimd:
OB = 3. 1ft. 00 = 0.96 ft. Oi/ = 1.72 ft.
286 MOTION OF RIGID BODIES
The velocity of B is,
vb = o) XAB = 10 Xt^ = 5 ft./sec.
Therefore the angular velocity of B, considered as a point on BC, is,
t03 = ^ = ^ = 1 . 61 rad./sec.
and the angular velocities of all points on BC are the same. The linear velocity
of C is, therefore,
yc = co3XOC = 1.61X0. 96 = 1.54 ft./sec.
And,
?'// = a,3XOfl' = 1.6lXl.72=2.77ft./sec.
Therefore,
yc 1.54 „ „_ , ,
a>2 = yrpj = —^ = 0 . 77 rad. /sec.
PROBLEMS
338. What is the magnitude and the direction of the velocity of the
point E (Fig. 322) on link DC midway between D and C.
Ans. ve = 0.77 ft./sec. perpendicular to DC.
339. Consider the link BC (Fig. 322) to be the triangular piece BOC and
let P be a point on the triangular piece midway from 0 to H. What is the
magnitude and the direction of the velocity of the point P?
340. Locate the instantaneous center of the connecting rod of the steam-
engine mechanism described in Problem 334. Find the angular velocity of
the connecting rod, the linear velocity of the crosshead, and the linear velocity
of a point on the connecting rod midway from the crosshead to the crank-pin.
Ans. (02 = 5.47 rad./sec; vp = 7.70 ft./sec; vm = ^A2 ft./sec.
341. Find the instantaneous center of the wheel described in Problem 335.
Find also the angular velocity of the wheel and the magnitudes and directions
of the velocities of the points B, C, and D.
PART III. KINETICS
CHAPTER IX
FORCE, MASS, AND ACCELERATION
§ 1. Preliminary Considerations. Kinetics of a Particle
136. Introduction. — Kinetics is that branch of mechanics
which treats of the laws in accordance with which the motion of
physical bodies takes place.
A change in the state of motion of a body always occurs when
an unbalanced force system acts on the body, the unbalanced
part (resultant) of the force system being the cause of the change
in the motion. Common experience teaches that the change
of motion of the body, is influenced both by the characteristics
(Art. 5) of the resultant of the forces acting on the body and by
the nature of the body itself. For example, different force systems
acting, in turn, on the same body do not produce the same change
of motion of the body. Further, the same force system applied
to different bodies, even if they are of the same size and form, does
not produce the same change in the motion of all of the bodies.
Although experience suggests that relations exist between
the force system acting on the body, the properties of the body,
and the change in the motion of the body, it required the work of
many eminent men and a period of several centuries before definite
and complete fundamental relations between these three factors
were finally established. These relations were formulated by
Sir Isaac Newton (1642-1727) and are known as Newton's laws
of motion.
Newton's fundamental laws, however, apply only to the
motion of a particle under the action of a single force, whereas
in engineering practice the motion of a body (system of par-
ticles) under the action of a system of forces must be con-
sidered. The force system acting on the body may produce any
287
288 FORCE, MASS, AND ACCELERATION
type of motion. Thus, in some cases a motion of translation is
given to the body, in other cases the body is given a motion of
rotation, and in still other cases the force system produces a plane
motion of the body. These three types of motion, only, are con-
sidered in this chapter.
137. The General Kinetics Problem. — In each type of motion
however, the general character of the kinetics problem is the same :
A physical body is acted on by a force system that has a resultant,
which causes a change in the motion of the body. In each prob-
lem it is required to deduce, by the use of Newton's laws, the
equations expressing the definite relations between (1) the result-
ant of the force system, (2) the kinetic properties of the body,
and (3) the change of motion of the body, so that the motion of a
given body produced by a given force system may be determined,
or the force system required to produce a given motion may be
found. The equations which express these relations are called
the equations of motion for the body.
These three elements or factors which are involved in the equa-
tions of motion of bodies may be considered briefly before stating
Newton's laws and before deriving the equations which express the
relations between these three factors. Change of motion is
measured in terms of distance, time, velocity, and acceleration,
relations between which, for various types of motion, have already
been considered fully in Chapters VII and VIIL The charac-
teristics of the resultant of a force system have also been consid-
ered in Chapter II, and need only be reviewed briefly at this point
(see next article) to show their connection with the general problem
in kinetics. The nature or property of the body which enables it
to have an influence in determining its own motion, however,
needs to be discussed at greater length (see Art. 139).
138. Characteristics of a Force System. — The only part of a
force system that influences the motion of a body is the unbal-
anced part (resultant) of the force system. The forces which
produce the 1 ypes of motion considered in this chapter are coplanar
forces or may be so regarded.^ The resultant of a coplanar force
^ Although in most kinetics problems here considered, the motions are pro-
duced by coplanar force systems, the system of forces may be non-coplanar.
It will be evident, however, after discussing the equations of motions for the
three types of motion here considered, that non-coplanar forces that pro-
duce these motions may be replaced by equivalent coplanar force systems.
INERTIA AND MASS 289
system is either a force or a couple (Art. 28) , although the resultant,
when a force, may, for convenience, be considered as another force
and a couple (Art. 18). If the resultant is a force, the character-
istics of the resultant which influence the motion of the body on
which the force system acts are (1) its magnitude, (2) the position
of its action line in the body, and (3) its sense (Art. 5). If, how-
ever, the resultant is a couple, the characteristics of the resultant
which influence the motion of the body on which the force system
acts are (1) the moment of the couple, (2) the sense, and (3) the
aspect or direction of the plane of the couple (Art. 16). These
statements appeal directly to one's experience and may be verified
by simple experiments which the reader may readily perform.
Since the equations of motion of a body must take account of
all the factors influencing the motion, there must be a sufficient
number of equations to determine the influence of all of the charac-
teristics of the resultant of the force system as mentioned above.
This may be done, for the types of motion considered in this
chapter, by means of three equations. These three equations will
contain the algebraic sum of the a;-components of the forces acting
on the body, the algebraic sum of the ^/-components, and the alge-
braic sum of the moments of the forces about some axis in the body.
For, if the resultant of the force system acting on the body is a
force, the algebraic sums of the x- and ^/-components of the forces
are needed to take account of the influence of the magnitude of
the resultant force and of the direction of its action line, and
the algebraic sum of the moments of the forces is needed to measure
the influence of the position of the action line of the resultant force.
And, if the resultant of the force system is a couple, the motion
of the body is one of rotation as will be seen later (Art. 147), in
which case, the algebraic sum of the moments of the forces is
needed to measure the influence of the moment and sense of the
couple. The aspect of the couple is not involved in the equations
of motion since the plane of the couple always agrees with that of
the plane of motion of the body.
139. Inertia and Mass. — The property of a body by virtue of
which it offers resistance to any change in its motion, and thereby
makes the body itself a factor in determining the motion which
unbalanced forces impress upon it, is called inertia. All physi-
cal bodies are inert or possess inertia, but different bodies possess
different amounts of inertia. That is, all bodies influence their
290 FORCE, MASS, AND ACCELERATION
own motion according to the same law but not to the same
degree.^
Bodies possess many otner properties such as hardness, grav-
itation, elasticity, strength, etc. In taking into account the
influence which any one of these properties has on the behavior
of the body, under the conditions which make the property a
factor in the problem, a quantity (value of the property) is found
which measures the particular property; that is, which indicates
its relation to the other influencing factors and which, if possible,
has a constant value for the body or for the material of the body.
Thus, when the property of strength is involved in a problem, the
constant values called " elastic limit " and " ultimate strength "
are frequently used as a measure of the strength of the material.
Similarly, the constant value called " modulus of elasticity "
is used as a measure of the property of elasticity or stiffness of a
material.
In the kinetics problem, that is, in a problem in which a change
in the motion of a body occurs, the property of inertia of the body
is involved. In taking account of the influence which the prop-
erty of inertia has on the motion of the body, a quantity called the
mass of the body is used. The two essential characteristics of
mass are its direct relation to the other factors (forces and accel-
eration) which are involved in the kinetics problem and its con-
stant value. Thus the mass of a body is a constant of the body
which measures a kinetic property (inertia) of the body.
The term mass then, as used in mechanics, denotes the amount
of inertia possessed by a body. Or, the mass of a body may be
defined as the constant of the body which measures the influence
which the body exerts in determining its own motion, when acted
upon by an unbalanced force system. The amount of resistance
offered by a body to a change in its motion (that is, the amount of
inertia possessed by a body), must be determined experimentally.
An experiment for determining the mass may be outlined as follows:
^ The way in which bodies resist motion is analogous to the manner in
which elastic materials, such as steel or timber, resist being stretched. Each
material, within limits, resists according to the same law (stretch is propor-
tional to stress), but some materials resist to a greater degree, that is, some
materials are stiffer than others, which means that some materials require a
greater force to produce a given stretch than do other materials, just as some
bodies require a greater force to overcome their inertia at a definite rate than
do other bodies.
INERTIA AND MASS 291
Let a body A be acted upon by a single force Fi causing a change
in its motion such that the acceleration of the body is found to be ai.
In the same manner let the forces F2, F3, etc., act on the same
body, in turn, causing accelerations which are found to be a2, as,
etc., respectively. The results of such an experiment show that,
Ml -t tl Q /T O
— = — = — = etc. = a constant = Ci (say) .
ai a2 as
If any other body is acted upon in a similar manner ^ by other
forces {F\ F" , F'" ^ etc.), one at a time, such that the body is given
the same series of accelerations (ai, a2, as, etc.) as was given to the
first body, it will be found that,
Y' F" F'"
— = — = = etc. = a constant = C2 (say) .
ai a2 as
That is, the manner in which a body influences its own motion
is by requiring that any force which is impressed on it shall bear
a constant ratio to the acceleration which the force gives to the
body. Further, it will be noted that the particular change in t*he
motion of the body that is related directly to the forces acting
on the body and the kinetic properties of the body is the accelera-
tion of the body.
Although the ratio of the impressed force to the acceleration
produced is constant for any given body it has different values for
different bodies. That is, C\ and C2 as used above are, as a rule,
not equal. Now C\ and C2 are measures of the inertia resistance
offered by the particular bodies to a change in their motion since
they measure the force required to overcome the inertia at a
definite rate (acceleration). The constants C\ and C2, therefore,
are proportional, respectively, to the masses of the two bodies.
That is, C\ = kMi and C2 = kM^ in which /c is a constant and M\
and M2 are the masses of the bodies. If now the mass, M2, of the
second body is taken as the unit by which the mass of the first body
is measured, then the ratio (C1/C2) of the constants expresses the
number of units of mass in the first body. The numerical value
of this ratio, therefore, depends upon the choice of the value of
C2, and hence, of the unit of mass.
^ It is assumed in this experiment that a motion of rectihnear translation
is given to the body in each case and hence the only characteristic of the force
that changes is its magnitude.
292 FORCE, MASS, AND ACCELERATION
F
Now, as shown above, C2 is expressed by the equation C2 = — .
a
Therefore, the unit of mass will depend upon the units in which
F and a are expressed. The various units of force and accelera-
tion, and the resulting units of mass are discussed in detail in
Art. 141. Mass is considered at this point only to show its general
connection with the other quantities (force and acceleration)
involved in the kinetics problem and to show a method of deter-
mining it.
The experiment suggested above for determining the mass of a
body by finding the ratio of an impressed force to the acceleration
produced, is difficult and inconvenient to perform. Fortunately,
a much simpler experiment serves the same purpose. Careful
experiments have shown that the weights (earth-pulls) of bodies,
when the weights (W) are the only forces acting on the bodies, pro-
duce the same acceleration, g, of all the bodies at a given location
on the earth's surface, the approximate value of g for most localities
being 32.2 ft. per sec. per sec. Hence, if instead of using the ratio
of any force, F, to the corresponding acceleration, a, the ratio
W
— is used, the only experiment needed to determine the mass of
Q
a body is that of weighing the body on an ordinary spring balance
since the value of g is known. Therefore,
a g'
But, as shown above, the ratio of the force F to the accelera-
tion, a, produced by the force is proportional to the mass ikf of the
F
body, that is, — = kM. Therefore,
W
— = kM.
9
And if units for W, g, and M are so chosen that k equals unity, as
will be discussed in Art. 141, then the number of units of mass in
a body may be found from its weight by the equation
M= — .
9
NEWTON'S LAWS 293
140. Newton's Laws. — Newton established his laws of motion
from a study of the motion of planets whose orbits and speeds
were then well known. Since the dimensions of a planet are
negligible in comparison with the range of its motion, Newton's
laws really apply directly only to a particle, that is, to a body, all
points of which may be considered at any instant to have the same
acceleration. In most cases of motion of bodies, however, the
accelerations of different particles of the body are not the same.
Thus, the accelerations of various parts of a rotating flywheel or
of the connecting rod of a steam engine are quite different. The
expression '' acceleration of a body " is, therefore, indefinite and
meaningless, except for a body having the motion of translation.
Newton's laws may be stated as follows:
First Law. — A particle remains at rest or continues to move
uniformly in a straight line unless acted upon by an unbalanced
force.
Second Law. — When a single (unbalanced) force acts on a par-
ticle the particle is accelerated; the direction of the acceleration
is the same as that of the force and its magnitude is directly
proportional to the force and inversely proportional to the mass
of the particle.
Third Law. — There are mutual actions between any two par-
ticles of a system (body) such that the action of the one on the
other is equal, collinear, and opposite to that of the other on the
one.
1. The first law is a qualitative one. It states that a particle
has inertia, that is, it resists having its motion changed. It states
that a force must act on the particle if its motion (velocity) is
changed either in direction or in magnitude, that is, if an accel-
eration is produced.
2. The second law is a quantitative one. It states what the
magnitude and the direction of the force must be in order to pro-
duce a given acceleration of a given particle, and shows that
although a particle cannot, of itself, change its state of motion it
does nevertheless influence the change of motion caused by the
force, by regulating or governing the manner in which the accelera-
tion shall take place; namely, that it shall be always inversely
proportional to the mass of the particle.
3. In the first two laws it is assumed that a single particle is
acted upon by a single force. But the third law brings out the
294 FORCE, MASS, AND ACCELERATION
fact that a single force does not exist. One force requires the pres-
ence of another force. A push cannot be exerted unless there is a
resisting force to develop the push; that is, forces always occur in
pairs. A pair of forces, however, requires the presence of a pair of
particles each of which acts on the other, and hence requires that
each offers a resistance to the other. The resistance which the
second particle offers to the first, in general, may arise in two
ways: (1) by transmitting to the first particle the force (static
reaction) which a third particle (or several particles) exerts on
it when holding it in equilibrium, or (2) by virtue of its inertia
as it allows the first particle to change its motion (kinetic reaction).
Or, the resistance may be a combination of these causes.
In a moving system of particles, therefore, the motion of each
particle is influenced by every other particle in the system. These
actions and reactions which influence the motion of the particles of
the body are, in general, impossible to determine. However, since
they occur as collinear pairs (third law) the equations of motion
for a system of particles (body) may be found without introducing
them into the equations, for, in finding the sums of x- and t/-com-
ponents and the sum of the moments of all the forces acting on all
the particles, the mutual actions and reactions between the pairs
of particles drop out of the expressions, since these sums are equal
to zero. Hence, by use of Newton's third law, the second law,
which appUes directly only to a single particle under the action of a
single force, may be used to extend the equations of motion to a
system of particles under the action of a system of forces.
141. Mathematical Statement of Newton's Second Law. Units.
— Newton's second law may be expressed mathematically by the
equation
F — kma,
in which a is the acceleration of the particle of mass m, F is the
single force acting on the particle, and A; is a constant factor the
value of which depends upon the units used to express the other
quantities {F, m, and a) in the equation. But in Art. 139 it was
F W
shown that — = — . Hence for any particle,
F W ,
— = — = km.
a g
NEWTON'S SECOND LAW. UNITS 295
It will be observed, therefore, that although the mass, m, of
F
a particle is always proportional to — , the numerical value of m
0/
(that is, the number of units of mass in the particle), is expressed
F W
by — (or more conveniently by — as discussed in Art. 139), only
d Q
when units are chosen for the force F and the acceleration a such
that k becomes unity. It is obvious that such a system of units
is desirable for, when such a system is chosen, Newton's second law
may be written simply
F = ma = — a.
9
A system of units which gives a value of unity to h may be
chosen as follows: In the equation F = kma four quantities are
involved. If a particular value (unity) is assigned to k then the
units for only two of the other quantities may be chosen arbitrarily.
In engineering, the unit of force and of acceleration are chosen
and hence the unit of mass is derived from the units selected for
force and acceleration. If the pound is chosen as the unit of force
and the foot per sec. per sec. as the unit of acceleration, as is
usual in engineering, it is evident that the body which has a unit
of mass is one which, when acted on by a unit force (pound), will
be given a unit acceleration (foot per second^). That is, if in the
equation F = kma, k is made unity, F is made unity (one pound),
and a is made unity (one foot per second^) , then m must be unity
(one unit).
The body which is given an acceleration of 1 ft. per sec. per sec.
by a 1-lb. force, that is, a body of unit mass, may be found by experi-
ment. The direct method would be to allow a 1-lb. force to act
on different bodies until one is found which is given an accelera-
tion of 1 ft. per sec. per sec. However, experiments have been
made which supply sufficient data from which the weight of a body
of unit mass may be found. Thus, experiments have shown that
the earth-pull on any body, if it is the only force acting, gives the
body an acceleration of gr, the value of which at most localities is
32.2 ft./sec.2 (approximately). Hence, if the earth-pull on a body
is 1 lb., that is, if the body weighs 1 lb., it will be given an accelera-
tion of 32.2 ft./sec.2 ^^^ ^^^ I ft./sec.^ And, if a 1-lb. force acts
on, a free body which weighs more than 1 lb. the acceleration pro-
duced will decrease in proportion to the increase in weight (New-
296 FORCE, MASS, AND ACCELERATION
ton's second law). Therefore, a body weighing 32.2 lb. will be
given an acceleration of 1 ft. per sec. per sec. by a 1-lb. force.
Therefore, a body weighing 32.2 lb. is a body of unit mass. A
body weighing 64.4 lb. has two units of mass, or, in general, a
W
body weighing W pounds has — units of mass.
No name for the unit of mass as here used has gained general
acceptance as has the name pound for the unit of force. The
name slug (from sluggishness which suggests inertia) is used to
some extent, particularly in England. The name geepound which
suggests 32.2 lb. is also used to sojne extent, but, in general, the
unit of mass here considered is called simply a unit of mass (some-
times the engineer's unit of mass as contrasted with that used by the
physicist). If a special name is given to it, such as a slug or gee-
pound, the tendency is to think of the unit of mass as an arbi-
trarily chosen unit like that of force and of acceleration and to
forget that it is a derived unit; derived from the units of force and
acceleration. Thus, the number of units of mass in a body when
W .
expressed by — , as discussed above, will be a certain number of
lb sec
— ^77 — '-, for, according to the dimensional equation for mass,
It.
Force F FT^
mass =
Acceleration L L
Hence the mass of a body which weighs 966 lb. is
W_ 966 ^o(.lb. sec.2
g "32.2 "^ ft. •
Or, if each of the units of mass is called a slug or a geepound the
mass of the body may be denoted as 30 slugs or 30 geepounds. A
lb sec
quantity expressed in the units —^ — - is difficult to visualize
or to interpret as a physical quantity, which accounts for the fact
that mass as used by the engineer sometimes fails to appeal to
common experience.
142. Other Systems of Units. — If the unit of force and unit
of acceleration are chosen arbitrarily and the unit of mass derived
therefrom, as was done in the preceding article, the system of
units thus obtained is called a gravitotional system.
OTHER SYSTEMS OF UNITS 297
If the unit of mass and of acceleration are chosen arbitrarily
and the unit of force derived therefrom, the system of units thus
obtained is called an absolute system.
If, in either the gravitational or the absolute system of units,
the units are so chosen that the value of k in the equation F = kma
becomes unity, the system is called a systematic or kinetic system.
If the kilogram is selected as the unit of force and the meter
per sec. per sec. as the unit of acceleration instead of selecting
the pound and the foot per sec. per sec. as was done in the preced-
ing article, then the number of units of mass in the body will be,
W
as before, expressed by — , but each unit of the mass will be ex-
pressed in terms of kilograms, meters, and seconds, instead of
pounds, feet, and seconds.
A systematic absolute system of units commonly used by physi-
cists in accurate scientific work and in engineering electrical meas-
urements is a system in which the gram, centimeter, and second
are used as the units of mass, length, and time, respectively.
Thus, if the gram is chosen for the unit of mass and the centimeter
per sec. per sec. for the unit of acceleration and if k is to have a
value of unity, then in the equation F = kma, F must be unity.
Hence, the force that will give a mass of 1 gram an acceleration of
1 cm. per sec. per sec. is the unit of force and is called a dyne.
A dyne is -q\^ of a gram. This fact may be shown as follows:
The earth-pull on a body of unit mass (gram) produces an acceler-
ation of 981 cm. per sec. per sec. instead of 1 cm. per sec. per sec.
Hence, from Newton's law, it follows that a force of ^^ gram
acting on the same body (unit mass) will produce an acceleration
of 1 cm. per sec. per sec. That is, ^-|^ gram is the unit of force
in the absolute system of units.
If the pound is used for the unit of mass and the foot per sec.
per sec. for the unit of acceleration, an absolute system of units is
obtained in which the unit of force is called the poundal, corre-
sponding to the dyne in the metric absolute system. The poundal
is 3272 of a pound. This fact may be shown by reasoning in a way
similar to that used above in showing that a dyne is -^^ of a
gram.
143. Equations of Motion for a Particle. — Since Newton's
laws apply directly only to a particle, they are limited to the action
of a single force or of a force system which has a single force (not a
298 FORCE, MASS, AND ACCELERATION
couple) for a resultant, for the reason that the force system which
acts on a particle is a concurrent system having as its resultant a
single force acting through the particle (point of concurrence).
Newton's laws, therefore, state the relation between (1) the result-
ant force acting on a particle, (2) the mass of the particle, and (3)
the acceleration of the particle. As noted above, however, by
means of the third law, this relation may be extended so as to
apply to the motion of a system of particles under the action of a
system of forces.
From Newton's second law the equation of motion of a par-
ticle, when systematic units are used, is,
R=ma
in which R is the resultant of all the (concurrent) forces acting on
the particle, m is the mass of the particle, and a is the acceleration
of the particle. The action line of R passes through the particle
since the particle is the point of concurrence of the forces acting
on it. Both R and a are directed quantities and their directions
always agree; hence, both sides of the equation may be resolved
into components. For convenience, three rectangular axes,
X, y, and z, are chosen and, since Rx=^Fx, Ry=^Fy, etc., three
equations of motion may be written as follows :
^F, = ma^
in which SF^., XFy, l^F^ are the algebraic sums, in the specified
directions, of all the forces acting on the particle.
It is important to note that all of the forces acting on the
particle are external to the particle. But, when the particle
considered is one of a system of particles, some of the forces acting
on it are exerted by the other (neighboring) particles of the system
and are, therefore, internal forces when considering the motion of
the whole system of particles.
Note. — In the problems which follow, the assumption is m«xde that the
bodies having the motions described may be considered to be particles without
introducing serious errors in the analysis of the motion. The problems may
be divided roughly into two classes; (1) the forces acting on the particle are
given and the characteristics (distance, time, velocity, acceleration) of the
resulting motion are rec^uired, and (2) the characteristics of the motion are
EQUATIONS OF MOTIONS FOR A PARTICLE
299
given and the forces required to produce the motion are required. In solving
the problems, therefore, the fact should be kept in mind that the kinematics
equations of Chapter VII are to be used, when needed, in conjunction with the
equations of motion of the particle, since the acceleration of the particle is
involved both in the kinematics and the kinetics of the particle. In solving
problems in kinetics it is important to follow a rather definite procedure the
main steps in which may be outlined as foUows:
^ 1. Determine carefully what is given in the problem.
>• 2. Determine carefully what is required in the problem. These two
steps, as a rule, require a sketch with the known and unknown
quantities either indicated on the sketch or briefly listed.
3. Draw a complete free-body diagram of the body (particle). That
is, show the actions of all other bodies on the particle considered.
This diagram is of particular importance in determining what
shall be used for the left-hand member of each of the equations
of motion for the particle.
4. Write the three equations of motion for the particle and select the
three coordinate axes tp be used in appljdng the equations.
Frequently, by a proper choice of axes, only two or even one of
the equations of motion will be sufficient for the solution.
5. Write any necessary additional equations such as kinematics equa-
tions, definitions, equations of equilibrium, etc., which apply to
the particular problem, keeping in mind that there must be as
many equations as there are unknown quantities.
6. Solve the equations, using the known quantities stated in the prob-
lem.
ILLUSTRATIVE PROBLEMS
342. A box weighing 16.1 lb. rests on the floor of an elevator. If the ele-
vator starts up with an acceleration of 8 ft. per sec. 2, what is the pressure on
the floor of the elevator?
Solution. — The box is given the acceleration
of 8 ft. per sec. per sec. by the upward pressure,
P, of the elevator and the downward weight,
W, as indicated in the freebody diagram (Fig.
323). Taking the ?/-axis as vertical, we need
only one of the equations of motion, namely,
'SFp = may, for the solution of the problem.
Hence using this equation we hav"
P-W
w
-dy,
Therefore,
^-i^-i=i^x^'
P = 20.1 lb.
Fig. 323.
300
FORCE, MASS, AND ACCELERATION
343. A small body weighing 4 lb. is attached to one end of a string
5 ft. long and is made to revolve as a conical pendulum with a constant angular
velocity, co, so that the string is inclined 30° with the vertical as shown in
Fig. 324. What is the tension, T, in the string and the Unear velocity, y, of
the body?
FiQ. 324.
Solution. — The body moves on a circular path in a horizontal plane under
the influence of two forces T and W as shown in the free-body diagram (Fig.
324). The acceleration of the body is rio^, or — , toward the center of the circle.
The equations of motion are:
From (1),
From (3),
W , Wv^
W
XFt = mat= — ra = 0, since a = 0.
SF^ = m%=0, since %=0.
4 v^
32.2'^5sin30°"
T cos 30° -4 = 0
The value of T may be found from (5). Thus,
(1)
(2)
(3)
(4)
(5)
T =
cos 30° .866
4.621b.
EQUATIONS OF MOTION FOR A PARTICLE
301
By substituting this value of T in (4), the value of v may be found. Thus,
4.62 cos 60^
:Xi
Hence,
Therefore,
32.2''5sin30°"
, 4.62X0.5X32.2X5X0.5
,2 = _
y = 6.8 ft. /sec.
=46.3.
344. Boxes are sent from the street into the basement of a store by means
of an inclined plane as shown in Fig. 325. The plane is 20 ft. long and makes
an angle of 30° with the floor. The boxes are given an initial velocity, vo, of
10 ft. per sec. Assuming that the coefficient of friction for the box while on
the incline is 0.4, what is the velocity, ?;, of the box as it reaches the bottom
of the incline and how many seconds does it take to reach the bottom?
f'yyyyyyy/y/yy/y/yy?
Floor , __,,..„.,^
Fig. 325.
Solution. — The forces acting on the box during its motion down the inchne
are shown in Fig. 325. The a;-axis is chosen in the direction of the total accel-
eration, hence ax = a.
The equations of motion are,
i:Fx = max = ma, (1)
i:Fi, = may = 0, since ay = 0 (2)
In addition to the equations of motion the following equations are needed :
F = nN, (3)
v^ = vo''+2as, (4)
v = vo+at (5)
From (1),
From (2),
From (3),
Wsin30°-F=— a.
9
N-Wcos30°=0. .
F=0.4iV.
(6)
(7)
(8)
302
FORCE, MASS. AND ACCELERATION
Substituting the value of F from (8) and (7) in (6) we obtain,
W
0.51f-0.4X.866T^=^a.
Therefore,
And from (4)
Therefore,
And from (5),
Therefore,
a = 4.95ft./sec.2
y2 = (10)2+2X4.95X20.
y = 17.3ft./sec.
17.3-10+4.95«.
^ = 1.47 sec.
PROBLEMS
345. A balloon weighing 400 lb. has a vertical acceleration of 2 ft. per sec.^
The horizontal wind pressure causes the balloon to travel in a direction making
an angle of 30° with the vertical. Find the horizontal component of the accel-
eration of the balloon and the horizontal wind pressure.
Ans. 1.15 ft./sec.2; 14.35 1b.
346. The winding drum of a mine hoist (Fig. 326) is 15 ft. in diameter.
A cage weighing 8 tons is raised by it. If the cage is rising at the rate of 60
ft. per sec. when it is 104 ft. below the surface of the ground, and power is
then shut off, what is the tension, T, in the cable while the cage is coming
to the surface if it comes to rest just as it reaches the surface?
Ans. 7410 1b.
347. An engine weighing 64.4 tons travels round
a curve having a constant radius of 1000 ft., at a
constant speed of 45 mUes per hour. Find the hori-
zontal thrust on the rails.
. ^
>^'W = 60 ft./ sec.
Fig. 326.
Fig. 327.
348. In " looping the loop " (Fig. 327), show that if friction is neglected,
the minimum value of the velocity of the car when at C is -yJ^- if the car does
not leave the track.
INTRODUCTION. METHODS OF ANALYSIS 303
349. A small body weighing 12 lb. rests on an inclined surface (Fig. 328)
which is revolved about a vertical axis with a constant angular velocity of 20
r.p.m. If the body is attached to the axis of ro-
tation by a cord as shown in the figure, and if
friction between the body and plane is neglected,
find the tension, T, in the cord.
350. A box is projected up an inclined plane,
which makes an angle of 20° with the horizontal,
with an initial velocity of 2400 ft. per minute.
If the coefficient of kinetic friction is 0.2, how
far up the plane does the box travel before
coming to rest? Will the box remain at rest?
If not, how long does it take it to reach
the bottom of the incline? Fig. 328.
Ans. s = 46.9 ft.
§ 2. Kinetics of Bodies f
144. Introduction. Methods of Analysis. — As stated in the
preceding section, the general character of a problem in kinetics
of bodies is the same whether the body has a motion of translation,
of rotation, or of plane motion: A physical body is acted on by a
force system that has a resultant which causes a change in the
motion of the body. For each type of motion, relations are found
between (1) the resultant of the external force system, (2) the
properties of the body (mass, moment of inertia, etc.), and (3)
the change in the motion of the body. For each of the types of
motion of rigid bodies treated in this section, the equations which
express the relations between the three factors or elements in the
problem (equations of motion) are found by the same procedure
or series of steps, as follows :
1. From the motion of the body, the acceleration, a, of any
(and every) particle in the body is found, both in magnitude
and in direction. This step involves the use of the equations
of kinematics as developed in Chapters VII and VIII.
2. From the acceleration, a, of any particle and its mass, m,
the force required to produce the acceleration is found, both in
magnitude and in direction, by applying Newton's second law.
This force, R, is expressed by the equation R = ma, its direction
being the same as that of a. This force is called the effective force
for the particle and is, of course, the resultant of the actual forces
acting on the particle. Some (most) of the particles of a body
are acted on by internal forces only (in addition to their weights),
304 FORCE, MASS, AND ACCELERATION
that is, by the neighboring particles of the body, and some of
the particles are acted upon by both internal and external forces.
The effective force for a particle, therefore, may be the resultant
of internal forces only or of internal and external forces.
3. The magnitude and the direction of the effective force for
each particle of the body having been determined, in terms of the
mass and acceleration of the particle, the resultant of the effective
forces for all the particles of the body is found completely by the
same methods as were used in Chapter II for finding the resultant
of a given system of forces. The effective forces for bodies having
the motions considered in this section may be assumed to form a
coplanar force system.^ Therefore, the resultant of the effective
forces, in general, may be determined by writing three equations
(Art. 30) involving the summations of the a;-components of the
effectives forces, of the ^/-components of the effectiye forces, and
of the moments of the effective forces about an axis in the body.
4. The equations obtained in step 3, which determine the
resultant of the effective forces for the whole body, contain all
the internal forces and all the external forces which act on all of
the particles. But, in obtaining the summations of the x- and of
the i/-components and of the moments of the effective forces, the
internal forces drop out of the expressions since they occur in
collinear pairs, the forces of each pair being equal and opposite
(Newton's third law). Hence, the sum of their components in
any direction and the sum of their moments about any axis are
zero. Therefore,
The resultant of the effective forces for the particles of a
^ In the discussion of translation, in this section, the motion will be restricted
to a uniplanar translation. Further, the bodies considered in connection with
the three types of motion discussed in this chapter will be assumed to be
homogeneous and to have planes of symmetry which are the planes of motion
of the bodies (Arts. 133 and 134) and hence, in discussing the kinetics of such
a body having a motion of translation or rotation, or a plane motion, the mass
of the body may be assumed to be concentrated in the plane of motion. Thus,
the body may first be imagined to be made up of elementary rods perpendicular
to the plane of symmetry and then each rod may be imagined to be squeezed
or compressed without having its cross-sectional area changed, until the rod
becomes an elemental disc lying in the plane of symmetry and having the same
mass as does the elementary rod. Therefore, the effective forces may be con-
sidered to form a coplanar force system in the plane of motion of the body.
If the body does not have a plane of symmetry, more than three equations are
needed to define its motion completely.
KINETICS OF A TRANSLATING RIGID BODY 305
body is identical with the resultant of the external forces
which act on the body. Or,
The resultant of the effective forces for all the particles
of a body, if reversed and assumed to act on the body with
the external forces, will hold the body in equilibrium.
This is a statement of D' Alembert' s principle. It will be noted,
therefore, that D'Alembert's principle makes it possible to reduce
a problem in kinetics to an equivalent problem in statics by intro-
ducing a force (or forces) which may be found completely from the
motion of the body by means of the first three steps outlined above.
Translation
145. Kinetics of a Translating Rigid Body. — If a rigid body is
acted on by an unbalanced force system that causes a motion of
translation, the resulting accelerations of all the particles of the
body are the same, both in magnitude and in direction (Art. 182).
And, whenever a particle of a body is accelerated, a resultant force
must act on the particle. This force, as already noted, is called
the effective force for the particle. The magnitude of the effective
force is the product of the mass, m, of the particle and its accelera-
tion, a, and its direction is the same as that of a (R = ma from
Newton's second law). Therefore, the effective forces for the
particles of a translating rigid body form a parallel system and,
since all of the forces have the same sense, their resultant must be a
single force (never a couple), the direction of which is the same as
that of each of the effective forces. And, as noted above, if the
body is symmetrical with respect to a plane through the mass-
center (plane of motion), the parallel forces may be treated as a
coplanar parallel system (see footnote under Art. 144). Thus,
let the rigid body represented in Fig. 329(a) be given a rectilinear
translation along a smooth plane by the force P which produces
the acceleration, a, of each particle of the body. Let the body
be made up of sixteen small cubes (representing the particles
of the body) attached (glued) so that they move as a rigid system.
The cubes may be made of different materials and hence their
masses are not necessarily equal. For convenience, each cube will
be thought of as a particle. The effective force for the first par-
ticle is mia, for the second particle, m2a, etc. The sixteen effect-
ive forces form a coplanar parallel system as shown.
306
FORCE, MASS, AND ACCELERATION
Resultant of Effective Forces. — The resultant of these effective
forces may be found in magnitude, action Une, and sense by the
method used in Art. 26 for finding the resultant of a sytem of
parallel forces in a plane. Thus, the resultant of the effective
forces is a force having a magnitude equal to the sum of the forces,
its sense being the same as that of the common acceleration.
Hence,
The magnitude of the resultant of the effective forces = 'Lma
= aXm
= Ma
V
WJ
\
r^"*'"
X
-*■
-*
X
m^a
\
^
-*
r*"
3
'^__r-direction^
-^
—
3
-►
—
->-
^
^/w/////)//////,
///.
^////////////////////
'W
(a)
N
X
(
)
\,
-^
-*
->
-►
lij
\
— *•
p
->
ii
-^
->
-►
1,
-^
->
-*■
-^
Ma
tw
(6)
N
Fig. 329.
in which M denotes the mass of the whole body, that is, the sum of
the masses of the particles {M=Xm) and a is the acceleration,
which is common to all the particles and hence is a constant and
may, therefore, be taken outside the summation sign.
The action hne of the resultant of the effective forces is located
by use of the principle of moments (Art. 25). Thus, in Fig. 329(6),
the sum of the moments of the effective (ma) forces about any point,
0, must equal the moment of their resultant (Ma) about the same
point. Hence, if y denotes the moment-arm of the effective force
for any particle with respect to 0, and p denotes the moment arm
of the resultant Ma with respect to 0, the principle of moments is
expressed by the equation,
That is,
Therefore,
X{'ma-y)=Ma-p.
X{my)=Mp, since a is constant.
S(m^
^~ M '
KINETICS OF A TRANSLATING RIGID BODY 307
This equation shows, however, that the resultant of the effective
forces passes through the mass-center (and not as shown in Fig.
3296) since the expression for p is also the expression for the y-coor-
dinate (y) of the mass-center (Art. 85).
Relation between Effective Forces and External Forces. — The
relation between the resultant of the effective forces and the
resultant of the external forces which act on the body may now be
found. The resultant of the effective forces is the resultant of
all the forces acting on all the particles and hence includes all of
the internal forces and all of the external forces. Therefore,
if Fr denotes the component of a force parallel to the direction of
a, we may write,
i^Fr) External4- (2F,) internal = ^(ma) = Ma.
However, in obtaining the summation of the r-components of all
the effective forces, the sum of the r-components of the internal
forces becomes zero since the internal forces occur in pairs of equal,
opposite, and collinear forces (Newton's third law). Therefore,
the resultant of the external forces is identical with the resultant
of the effective forces (D'Alembert's principle). That is, it has the
same magnitude, action line, and sense as the resultant of the
effective forces. Hence, the magnitude of the resultant of the
external forces (2/^r) is given by the equation
i:Fr = Ma.
Thus, the resultant of the external forces (P, TF, and N) is
represented completely by the force Ma in Fig. 330(a). Or, to
express the same idea in other words, if an additional force were
acting on the body with the external forces P, W, and N, the mag-
nitude and the action line of the additional force being the same as
the magnitude and the action hne of the resultant (Ma) of the
effective forces, but reversed in sense (opposite to a), as shown in
Fig. 330(6), then the forces acting on the body would be in
equilibrium. This additional (imaginary) force is sometimes
called the reversed effective force for the body or the inertia force
for the body. It will be noted that the normal pressure N cannot
be collinear with W when the force P has the action line as given
in Fig. 330(6).
308
FORCE, MASS, AND ACCELERATION
The fact that the action Une of the resultant of the external
forces passes through the mass-center of the body may be expressed
by stating that the sum of the moments of the external forces with
respect to an axis through the mass-center is zero (ST' = 0).
^yyyy/y//////yk;c
Ma
fW(a)
i^///A//yyzy///;^/A '^y/M^/y/yy/y//
W
Ma-
Fig. 330.
Therefore, the equations which express the relations between
the external forces, the kinetic property of the body (mass), and
the change of motion (acceleration) for a translating rigid body are
ST = 0
(1)
For convenience, the first of these two equations may be replaced
by two equations which involve the components of the external
forces and of the acceleration of the body, in the directions of two
rectangular axes. Thus if x and y denote the rectangular axes,
the equations of motion then become.
l.F, = Ma,
^Fy = May
SJ = 0
(2)
If the X-axis is chosen in the direction of the acceleration, then
ax is the total acceleration. The second equation then becomes
ItFy = 0 since ay is zero, that is, the body is in equilibrium as far as
motion in a direction perpendicular to the direction of the accelera-
tion is concerned, and equations (2) reduce to equations (1).
However, the equation 2jPy = 0 may give an important relation
KINETICS OF A TRANSLATING RIGID BODY
309
between the forces in the solution of a particular problem. Further,
it is sometimes convenient to select x- and i/-axes neither of which
agree in direction with the acceleration. Equations (2) therefore,
will, as a rule, be used and referred to as the equations of motion
of a translating rigid body.
Note. — In analyzing and solving problems in kinetics of bodies, the same
general procedure should be followed as was outlined in the note at the end
of Art. 143.
^^
ILLUSTRATIVE PROBLEMS
361. The dimensions of block A (Fig. 331) are 3 ft. by 3 ft. by 5 ft. and the
weight of the block is 1200 lb. The block rests on a carriage, B, which is given
an acceleration a in the direction shown.
If the friction between the block and car-
riage is sufficient to prevent slipping,
what is the maximum acceleration that
the carriage can have without causing the
block to tip over?
Solution. — The block has a motion of
translation under the action of two forces,
namely, the weight, W, and the reaction,
R, of the carriage. For convenience
the latter force, which acts at O when
the block is on the point of tipping, will
be resolved into the normal pressure, A^,
and the frictional force, F, as indicated in
the figure. The equations of motion for Fig. 331.
the block are,
i:F:, = Max, (1)
W=^
B
^^^^^^^^^^^^
XFy^Mny,
2r = 0.
(2)
(3)
Since the rr-axis is chosen in the direction of the acceleration of the body, it
follows that ax = a and ay = 0.
From (1),
1200
F = a (4)
32.2 ^
From (2),
iV-1200 = 0, (5)
From (3),
fF-fiV = 0, (6)
310
FORCE, MASS, AND ACCELERATION
By solving (5) and (6) for F and substituting its value in (4), the values of F
and a become,
F = 72Q\h.,
a = 19.32ft./sec.2
Second Method. Inertia Force Method. — If the inertia force (reversed
effective force) for the body is assumed to act on the body with the external
forces, the body may be assumed to be in equilibrium (D'Alembert's principle)
and hence the equations of equilibrium may be applied to the force system thus
formed.
The inertia force for the translating block A is,
AT 1200
Its direction is opposite to that of a and its action line passes through the mass-
center of the block. Therefore, the forces acting on the block as shown in
120 lb.
Fig. 332.
Fig. 333.
Fig. 332 will hold the block in equilibrium. The equations of equilibrium
for the force system (Art. 50) are,
' 1200^ .
^^^ = ^"3272'^=^'
7:Fy=N -1200 = 0,
2Mo=|50ax|-1200x| = 0.
The solution of the equations leads to the same results as were found by
the first method of solution.
It should be noted that in obtaining the moments of the forces in the above
equilibrium equation (SMo = 0), the moment-center O may be taken as any
point in the plane of the forces, whereas the moments of the forces in the third
equation of motion {ZT = 0) used in the first method of solution must be taken
about the mass-center of the body.
362. Three bodies A, B, and C are connected by two cords as shown in
Fig, 333 so that all three bodies move with the same acceleration. The pulley,
KINETICS OF A TRANSLATING RIGID BODY
311
D, is assumed to be weightless and frictionless. Body A weighs 36 lb., B
weighs 60 lb., and C weighs 120 lb. The coefficient of kinetic friction for A
and the plane is j, and for B and the plane it is t. Find (1) the acceleration
of the bodies, (2) the tension, Ti, in the cord between A and B, and (3) the
tension, T2, in the cord between B and C.
Solution. — Each body has a motion of translation. Hence, the equations
of motion for a translating body apply to each body; namely.
i:Fr. = Maa
'LFy = May,
sr=o.
(1)
(2)
(3)
In addition, two equations involving the coefficients of friction for bodies
A and B are needed, namely,
FA=iNA, (4)
FB=iNB.
(5)
Equation (3) is not needed in determining the quantities asked for. If the
action lines of Na and A^^ were desired, however, all five equations would be
needed.
A free-body diagram for body A is shown in Fig. 334(a). Appljdng the
equations of motion to A we have,
£■
(c)
1201b.
Fig. 334.
From (1),
From (2),
From (4),
36
^1-36 sin 30° -/^A =^^r::0'x-
iVA-36cos30° = 0.
Fa = \Na.
(6)
(7)
(8)
By solving for Fa from (7) and (8) and substituting its value in (6), the
following equation is obtained.
Of!
ri-36sin30°-iX36cos30° = — — aa; (I)
oZ . Z
312
FORCE, MASS, AND ACCELERATION
In like manner, by applying the equations of motion to body B, the
following equations are obtained (see Fig. 334(5) for free-body diagram):
^2-7^1-^5-60 sin 30°= J5_a^,
iV£-60cos30°=0,
Fb=\Nb.
60
From the last three equations we obtain,
7^-7\-iX60 cos 30° -60 sin Zi)'' =-~—ax.
For the body C the equation of motion is (see Fig. 334c),
120
120-7^2
32.2
ax.
(II)
(III)
Equations I, II, and III contain three unknowns, T\, Ti, and ax or a (the
X-axis is chosen in each case in the direction of the total acceleration, a). By
solving the three equations for the three unknowns the following results are
obtained;
a = ax = 8.02ft./sec.2,
7^1 = 34.76 lb.,
7^2 = 90.1 lb.
353. An elevator starts from rest and moves upwards, acquiring a velocity
of 800 ft. per minute in a distance of 18 ft. If the acceleration is constant
what is the pressure of a man on the floor of the elevator if the man weighs
161 lb.?
Solution. — The man forms the body (assumed rigid), the motion of which is
considered. Since the body has a motion of translation, the three equations
of motion for a translating rigid body apply, but
since the forces acting on the body pass through
the mass-center and since one of the axes is chosen
in the direction of the acceleration, only one of the
equations is needed; namely,
(1)
In addition to the equation of motion one
kinematics equation is needed; namely.
■VQ^-\-2as.
(2)
Fig. 335 shows the free-body diagram, the two
forces being collinear.
From (1),
From (2),
^ ... 161 161
^-l^l=32:2^^ = 32-2«-
(^y=0^2.XlS.
KINETICS OF A TRANSLATING RIGID BODY
313
By solving the last two equations the following results are found,
a = 4.93ft./sec.2,
P = 185.7 lb.
In the above problems the bodies have rectilinear translation only. The
parallel or side rod of a locomotive when running on a straight track has a
motion of curvihnear translation with reference to the engine frame. Its
motion is considered in the following problem.
354. The parallel rod of a locomotive (Fig. 336a) weighs 400 lb. The
crank length, n, is 15 in. and the radius, r2, of the drivers is 3 ft. If the
speed of the engine is 50 miles per hour, what is the reaction of the pin at
each end of the rod when the rod is in its lowest position?
(a)
Mr^u)- 1= 9270 lb.
I X I
@
Ri
W=400 lb.
(t>)
Fig. 336.
55
R,
Solution. — Each particle of the rod has the same acceleration at any instant.
When the rod is in its lowest position, the acceleration of each particle with
reference to the engine frame is directed vertically upwards, its value being
(oVi. The angular velocity, co, is
V 5280X50^1 _. , . , .
"=r-r "60X60- ^3 =24-^^ ^^^-Z^^^'
The resultant of the effective forces acts through the mass-center and its
magnitude is,
400 1 ^
Ma = Ma,2ri=^X(24.44)2X^ = 9270 lb.
If this resultant is reversed and assumed to act on the body with the external
314
FORCE, MASS, AND ACCELERATION
forces, as shown in Fig. 336(6), the forces will be in equilibrium (D'Alembert's.
principle).
It will be observed that the forces form a parallel force system. The equa-
tions of equilibrium for a parallel force system (Art. 49) are.
Usmg (1),
Using (2),
Whence,
2F=0,
(1)
(2)
i2i+/22 -9270 -400 = 0.
i2iXZ-(9270+400)X-=0.
i2,=7?2 = 48351b.
PROBLEMS
365. A man who is just strong enough to lift a 150-lb. weight when standing
on the ground can lift a 200-lb, weight from the floor of an elevator when the
elevator is going down with a certain acceleration. What is the acceleration?
What weight can the man lift from the floor of the elevator when going up
with the same acceleration? -4ns. a = 8.05 ft. /sec. 2; TF = 1201b.
356. A 3-ton cage descending a shaft with a speed of 9 yd. per sec. is
brought to rest with a uniform acceleration in a distance of 18 ft. What is the
tension in the cable while the cage is coming to rest?
357. The dimensions of body A (Fig. 337) are 3 ft. by 2 ft. by 4 ft. and
its weight is 1000 lb. Assuming that the body will not slip on the carriage,
what is the maximum weight that B may have without causing A to tip over
when the acceleration of the carriage is 8 ft. per sec. 2? The pulley D is assumed
to be frictionless and weightless. An&. 201 lb.
t/sec. Z^y//Z^^,^y'A^///////A
< — z — >
■y/yy/yy/y/Z'/A
Fig. 337.
Fig. 338.
358. A man weighing 150 lb. leaves his room by way of a window which
is 50 ft. above the ground. He has a rope that is long enough to reach to
the ground but it can support a force only of 125 lb. What is the least velocity
with which he can reach the ground?
KINETICS OF A ROTATING RIGID BODY
315
359. Two strings pass over a smooth drum. On one side of the drum the
strings are attached to a 50-lb. weight; on the other side one string is attached
to a 40-lb. weight and the other string to a 30-lb, weight. Find the accelera-
tion of the weights and the tension in each of the strings during motion.
Ans. 7^30 = 25 lb.; ^40 = 33.3 lb.; a = 5.37 ft./sec.^
360. Steam is shut off when a train running at a speed of 30 miles per hour
reaches a 0.4 per cent down-grade. What will be the velocity of the train after
100 sec. if the train resistance is 10 lb. per ton?
361. A small car (Fig. 338) with its load weighs 800 lb. and the center of
gravity, G, of the total weight is 5 ft. from the track. A force P of 120 lb. is
applied to the car as shown.
Neglecting the friction on the [<<: 3' >|< 3'
track, find the acceleration of
the car and the reactions of the
track on each pair of wheels.
What would be the reactions on
the wheels if the force P acted
through the point G?
362. A door is hung on a
track as shown in Fig. 339. The
coefficient of friction for each of
the shoes (A and B) and the
track is j. The door weighs
300 lb. What force P is required
to give the door an acceleration of 4 ft. per sec. 2? Find the reactions of the shoes
on the track. How far will the door travel in 2 sec? Ans. P = 112 lb.
Fig. 339.
Rotation
146. Kinetics of a Rotating Rigid Body. — If a rigid body is
acted on by an unbalanced force system that produces a motion of
rotation of the body, the resulting linear accelerations of the
various particles of the body are not the same, but vary directly
as the distances of the particles from the axis of rotation (Art. 133).
The angular velocities and accelerations of all the particles, how-
ever, are the same, at any instant, and the linear acceleration of
any (and every) particle in the body may be expressed in terms
of the common angular velocity, co, and the angular acceleration, a.
Thus, in Fig. 340 is represented a physical body acted on by the
external forces P, W, and the reaction (not shown) of the axis at
0, which cause the body to rotate with an angular acceleration, a,
about an axis through 0. Each particle of the body moves on a cir-
cular path and the linear acceleration of any particle in the body
316
FORCE, MASS, AND ACCELERATION
has two components; a tangential component, a^ and a normal
component, an, (Art. 118) such that,
a^ = ra, directed tangent to the path of the particle, and
a„ = rco2=—, directed normal to the path of the particle, towards
the center of rotation.
in which co and a are the angular velocity and angular acceleration,
respectively, of the particle (and also of the body) at the instant,
V is the linear velocity of the particle at the instant, and r is
the distance of the particle from the axis of rotation. The accel-
eration components for several particles are shown in Fig. 340.
From the components of the acceleration of any particle of the
body, expressed above, the components of the resultant force
(effective force) required to produce these acceleration components
may now be found. Thus, if m denotes the mass of the particle,
then from Newton's second law,
mat = mra is the tangential component of the effective force, and
man = mroi^ is the normal component of the effective force, as
shown in Fig. 341.
Resultant of the Effective Forces. — The resultant of the effective
forces for the whole body may now be found. As already noted
KINETICS OF A ROTATING RIGID BODY 317
(see footnote under Art. 144), the effective forces form a coplanar
system in the plane of motion and, as indicated in Fig. 341, the force
system is non-concurrent. According to Art. 28, the resultant of
such a coplanar system is either a force or a couple. In either case
the resultant may be determined completely from the summa-
tion of the components of the forces along each of two rect-
angular axes and the summation of the moments of the forces
about some point in the plane of the forces. Thus, let the two
axes be the T- and A'-axes as shown in Fig. 341, where the A-axis
is drawn through the center of rotation, 0, and the mass-center of
the body. Let each of the two components, mra and mrw^, of the
effective force for any particle be resolved parallel to the A- and T-
axes respectively. Let 6 be the angle (measured clockwise from
the A'-axis) between the radius, r, to the particle and the A-axis.
Then, from Fig, 341, the component of the effective force parallel
to the A-axis is,
— mroP' cos d — mra sin 6,
and the component of the effective force parallel to the T-axis is,
mra cos d — mroi^ sin 6.
Hence,
The algebraic sum of the components of the effective forces
parallel to the A-axis = S( — mrco^ cos 6 — mra sin d)
= — co^Smr cos d—a'Zmr sin 0
= - Mr cJ^-0.
In the above equation 2mr cos 6 represents the moment of the
body with respect to the T-axis and hence may be replaced by Mr,
and l^mr sin 6 represents the moment of the body with respect to
the A-axis and hence is zero since the A^-axis passes through the
mass-center of the body. Further, the negative sign indicates
that Mfoy^ is directed from the mass-center towards the center of
rotation. In like manner,
The algebraic sum of the components of the effective forces
parallel to the T'-axis = S(mra; cos d — mroP sin d)
=aXmr cos d — co^'Emr sin 0
= Mra-0.
318 FORCE, MASS, AND ACCELERATION
The moment of the effective force for any particle about the
center of rotation, 0, is mra • r since the normal component, mrco^,
of the effective force for each particle passes through the axis of
rotation and has, therefore, no moment about the center 0. Hence,
taking moments about 0 :
The algebraic sum of the moments of the effective forces with
respect to the axis of rotation =S(mrQ;-r)
where lo denotes the moment of inertia of the body with respect
to the axis of rotation.
Since the sum of the components, in any direction, of the forces
of a system is equal to the component of their resultant in the same
direction, and since the sum of the moments of the forces about
any point is equal to the moment of their resultant about the
same point, the above results may be summarized as follows:
The component of the resultant of the
effective forces parallel to the iV-axis = Mfoi^.
The component of the resultant of the
effective forces parallel to the T-axis = Mra.
The moment of the resultant of the
effective forces about the center of rotation = 7oa.
If the resultant of the effective forces is a force, all of the
above expressions are required to determine it completely. If,
however, the resultant is a couple, the first two expressions reduce
to zero, in which case Joa expresses the moment of the resultant
couple, as will be discussed in greater detail later in this article.
Relation between Effective Forces and External Forces. — The
relation between the resultant of the effective forces for the whole
body and the resultant of the external forces which act on the body
may now be found. Since the effective force for a particle is the
resultant of all the forces (both internal and external) which act
on the particle, the effective forces for all the particles will include
all of the internal forces (exerted by the particles among them-
selves) and all of the external forces. Hence, the summation of
the A/^-components of the effective forces is the same as the summa-
KINETICS OF A ROTATING RIGID BODY 319
tion of the A^-components of the internal forces plus the summation
of the iV-components of the external forces, and a similar statement
applies to the ^-components. Therefore, if F denotes a force
and T the moment of a force, the relation between the above
expressions and the forces acting on the particles is expressed by
the following equations:
(SF„) External +(2/^„) internal = ^fCO^
{^Ft ) External + {^Ft ) internal = Mr a
O^Tq) External + ( S Tq ) Internal = IqOL .
But the algebraic sum of the components, in any direction, of
the internal forces and the algebraic sum of the moments, about
the axis of rotation, of the internal forces are zero, since the
internal forces occur in pairs of equal, colUnear, and opposite
forces (Newton's third law). Therefore, the resultant of the
effective forces for the whole body and the resultant of the exter-
nal forces, only, are identical (D'Alembert's principle).
Therefore, in the above equations, the expressions which involve
the internal forces drop out since as just stated,
(SFw) Internal = 0, (2i^t) Internal = 0, (Sio) Internal = 0.
Hence, if F is now used to denote an external force only, and T is
used to denote the moment of an external force, the equations
which express the relations between (1) the external forces, (2)
the kinetic properties of the body, and (3) the change of motion
(the equations of motion for a rotating rigid body) are,
^Ft = MfoL
(1)
It will be noted that rci)'^ and ra are the normal and tangential
accelerations, respectively, of the mass-center of the body, and
hence the first two of the above equations may be written 2/^„ = Mctn
and ^Ft = M'dt.
Further, if the body rotates about an axis through the mass-
center, that is, if the points 0 and G coincide, then the right-
320
FORCE, MASS, AND ACCELERATION
hand members of the first two equations are equal to zero since r
is equal to zero, and therefore the algebraic sums of the components
of the external forces in the two directions are equal to zero.
Hence, if a rigid body rotates about an axis through its mass-
center, the equation of motion for the body is,
in which ST is the algebraic sum of the moments of the external
forces acting on the body with respect to an axis passing through
the mass-center (axis of rotation) and / is the moment of inertia
of the body with respect to the same axis.
ILLUSTRATIVE PROBLEMS
363. The rod BCE (Fig. 342a) is made to oscillate by means of the crank
y4Z) and link DC. The members are connected by smooth pins at B, C, and D.
The rod BCE has a constant cross-section and weighs 16.1 lb. In the position
shown its angular velocity, co, is 60 r.p.m. and its angular acceleration, a,
is 40 rad./sec.2 (see Art. 134 for method of finding w and a). Find the pressure,
P, of link DC at C and the reaction, R, of the pin at B.
Fig. 342.
, Solution. — The rod rotates about the center B. The equations of motion
for the rod are,
i:f„ = m?c^' . . . (1)
XFt^Mra, (2)
7:Tb = Ib<^ (3)
KINETICS OF A ROTATING RIGID BODY
321
Referring to the free-body diagram for the rod (Fig. 3426) we obtain:
From (1),
D ii«i ono 16.1^20^ /60X2ir\ 2
/?„ + 16.1cos30 =32:2X12X1-60-) (^^
From (2),
P-Rt+m.lsm30° = '^~^X^X40 (5)
From (3),
(16.1 cos 60°+P) ?5=75 « = |xi|']x Q 'X40. . . . (6)
Solving (4) we obtain,
Rn = 18.9 lb.
Solving (6) we obtain,
P = 36.4 lb.
Substituting the value of P in (5) and solving for Rt we obtain,
Rt = n.l lb.
72 = V(18.9)2 + (ll. 1)2 = 21.9 lb.
Therefore,
364. In Fig. 343, CD represents a brake for regulating the descent of the
suspended body, A. B is the drum from which the cable attached to A un-
winds as A descends. The radius, n,
of the drum is 6 ft. The radius, ra, of
the brake wheel is 7 ft. The radius of
gyration, A:o, of the rotating parts (drum
and brake wheel) about the axis of ro-
tation is 4 ft. The rotating parts
weigh 2000 lb. and the body A weighs
1000 lb. The coefficient of brake fric-
tion is J. If friction on the axle of
the rotating parts is neglected, find the
acceleration, a, of the body A, the
tension, P, in the cable, and the hori-
zontal and vertical components, Ri and
Ri, of the axle reaction, assuming the
force at C to be 100 lb. (Consider the
cable to be flexible and neglect its
weight.)
Solution. — Three bodies are to be considered, (1) the brake CD which is in
equilibrium, (2) the drum and the brake wheel which have a motion of rotation,
and (3) body A which has a motion of translation. The free-body diagram
for each body is shown in Fig. 344. The brake is held in equilibrium by a
non-concurrent force system in a plane for which the equations of equilibrium
are:
S/^x=0, (1)
Fig. 343.
SF, = 0,
2Mz> = 0.
(2)
(3)
322
FORCE, MASS, AND ACCELERATION
Equation (3) only is needed in this problem since all of the forces acting on
the brake are not required.
The equations of motion for the drum and brake wheel are:
'EFt = Mra,
(4)
(5)
(6)
In addition to these equations the defining equation of the coefficient of
friction must be used, namely,
F=fiN (7)
D4
—N
Qi
I Qa
(bj
Fig. 344.
One equation of motion only is needed for body A, namely,
(8)
1000 lb.
Further, since the total acceleration of body A is in the
^/-direction and since it has the same magnitude as the
tangential acceleration, at, of a point on the circumference
of the drum, we may write,
(9)
ay — a = at= ria
Applying the equations we have:
From (3),
Whence,
From (7),
100X4.5-0.5A^ = 0.
AT = 900 lb.
F = iX900 = 2251b.
KINETICS OF A ROTATING RIGID BODY
323
From (4),
Whence,
From (6),
From (8),
/2i -900 = 0, since r = 0.
/2i = 900 1b.
2000
6P-225X7=||^X42X«.
1000-P=|^X%.
(10)
(11)
Substituting - from (9) for a in (10) and replacing ay in (11) by;a from (9), we
have,
6P-226X7 = |5?X16X|,
1000-P=f5„.
and.
And, from (5),
/?2-P -225 -2000=0, since r=0.
These last three equations contain the three required quantities. The solution
of the equations gives,
a = 12.54 ft./sec,
P = 609 1b.,
i2, = 2834 1b.
365. Two spherical balls are connected by a light, slender, rigid rod and
made to rotate in a horizontal plane about a vertical axis midway between the
balls as shown in Fig. 345. Each sphere ^
is 12 in. in diameter and weighs 64.4 lb.
What turning moment must a couple
have which gives the spheres an angu-
lar velocity of 30 r.p.m. in 4 sec,
starting from rest? If one of the two
forces of the couple is applied 9 in. from
the axis of rotation and the other force
is the reaction of the axis, what is the
magnitude of each force?
Solution. — Since the two spheres
have a motion of rotation the equations of motion are,
XFn = M?io^ = 0, since r=0,
i:Ft = M¥a = 0, ....
2ro = /o«
Fig. 345.
(1)
(2)
(3)
324
FORCE, MASS, AND ACCELERATION
Letting the moment of the couple be denoted by C and the mass of each
sphere by M we have, from (3),
XTo = C = ha
= 2(7+Md2)« = 2(fMr2+Md2)a
= 6.65a.
But, from definition,
w-coo 30X27r __. , .
«= ^ = gQ^^ =0.785 rad. /sec.
Therefore,
(7 = 6.65X0.785 = 5.23 Ib.-ft.
And since,
C = FX^,
then
FX
A- = 5.23
Whence,
F = 6.97 lb.
PROBLEMS
366. A sohd sphere 15 in. in diameter revolves with an angular velocity of
500 r.p.m. about a fixed axis which passes through its center. What force
acting tangent to its surface will stop the sphere in 5 sec. if friction on the
axis is neglected? The weight of the sphere is 500 lb. Ans. 40.6 lb.
t
367. A weight of 30 lb. is suspended from a sohd homogeneous cylinder by a
weightless cord which is wrapped around the cyhnder. The cylinder weighs
193.2 lb. and its radius is 18 in. Bearing friction is 18 lb. and the diameter of
the shaft on which the cylinder rotates is 4 in. If the suspended weight has an
initial velocity of 10 ft. per sec. downwards, what will be its velocity after it
has moved 10 ft.? What time is required to move the 10 ft.?
368. A body C weighing 10 lb.
rests upon a frame D (Fig. 346) which
rotates about a vertical axis AB.
When the frame is not rotating, the
tension in the spring, S, is 20 lb. If
the angular velocity of the frame is
30 r.p.m. and the friction under C is
neglected, what is the pressure against
the stop at El Ans. 10.8 lb.
--^WWW —
Fig. 346.
369. If the rocker arm BCE in
Prob. 363 carries on its outer end E a
spherical ball which weighs 4 lb., find the values of P, Ru and Rn, assuming
that the ball is small enough to be considered as a particle.
370. A block A (Fig. 347) weighs 257.6 lb. and rests on a plane inchned
30° with the horizontal. A flexible rope attached to the block passes over the
KINETICS OF A ROTATING RIGID BODY
325
disc, C, at the top of the plane and is attached to a suspended body, B, which
weighs 644 lb. C weighs 322 lb. and is 20 in. in diameter. The axle is 4 in.
in diameter and the axle
friction is 10 lb. The co-
efficient of friction for body
A is 0.1. Find (a) the
acceleration of B, (h) the
angular acceleration of the
disc, (c) the pull of the
rope on body B. Neglect
the material cut from the
disc for the shaft.
371. A constant turning
moment of 240,000 in.-lb.
turns a flywheel, the weight
of which is 3220 lb. The
diameter of the flywheel is 8 ft
per sec. to 80 rad. per sec
Fig. 347.
Its angular velocity is changed from 40 rad.
in 20 revolutions. What is the radius of gyration
of the flywheel with respect to the axis of rotation? Neglect axle friction.
372. The drum B (Fig. 348) is
rotating with an angular velocity
bi = 120 r.p.m. when the brake C is
applied. The drum is a soUd cylin-
der and has a radius of 10 in. Its
weight is 2000 lb. If the coefficient
of brake friction is 0.2, what force,
P, is required to reduce the an-
gular velocity of the drum to 30
r.p.m. in 3 sec? Neglect axle
friction. Ans. P = 203 lb.
373. Two bodies, A and B (Fig. 349) rest on a frame, E, which rotates
about an axis CD. A weighs 8 lb. and B weighs 24 lb. The tension in
Fig. 348.
)u;
Fig. 349.
326 FORCE, MASS, AND ACCELERATION
the spring, S, when the frame is not rotating is 30 lb. If the friction under
bodies A and B and the mass of the spring are neglected, what are the pressures
against the stops F and G when the frame revolves with an angular velocity
of 30 r.p.m.?
374. In Prob. 365, what is the tension in the rod?
147. Second Method of Analysis. Inertia Forces. — In some
problems dealing with the rotation of a rigid body under the
action of an unbalanced force system, it is convenient to assume
that the resultant of the effective forces is reversed and acts on the
body with the external forces, thereby forming a force system
that is in equilibrium (D'Alembert's principle), and thus reducing
the kinetics problem of a rotating body to an equivalent statics
problem. In the preceding articles it was shown that the resultant
of the effective forces for a rotating body is either a force or a
couple in the plane of motion of the body. The reversed resultant
force (or resultant couple) is called the inertia force (or inertia
couple) for the body. In using this method of analysis it is
important, therefore, to determine under what conditions of the
motion the resultant is a force and under what conditions it is a
couple. Further, if the external forces are to be put in equilibrium
by the addition of the anti-resultant of the effective forces, that is,
by the inertia force for the body, the action hue of the resultant
force must be determined. The facts stated in the following
proposition are of great importance in connection with the method
of solution as outlined above.
Proposition. — If a rigid body rotates about a fixed axis through
its mass-center, the resultant of the effective forces (and of the
external forces also) is a couple the moment of which is equal to
la, the product of the moment of inertia of the body about
the axis of rotation and the angular acceleration of the body. If
the body rotates about an axis which does not pass through the
mass center, the resultant of the effective forces is a single force
having: a magnitude equal to the product of the mass of the
body and the acceleration of the mass-center of the body (Ma);
a direction the same as that of the mass-center; and an action
line which does not pass through the mass-center of the body. For
convenience, however, this single force may be considered to be
equivalent to another force, having an action Une which does pass
through the mass-center, and a couple; the magnitude of the force
SECOND METHOD OF ANALYSIS 327
is equal to Ma and its direction is the same as that of a; and the
moment of the couple is equal to la.
The facts stated in the above proposition, together with certain
additional facts, are discussed and applied in the following pages,
in connection with the use of inertia forces. It was found in the
preceding article that the expressions Mra and Mfo)^ represented
the T' and iV-components, respectively, and that loa represented
the moment, of the resultant of the effective force system for the
rotating body. From these expressions, the resultant of the
effective forces for various conditions of motion may be determined
completely. Thus, it will be noted that if f is zero in these expres-
sions, that is, if the distance from the axis of rotation to the mass-
center of the body is zero, then each of the components, Mra and
Mfui^, of the resultant of the effective forces is zero. Hence, the
resultant is not a force. And, since the effective forces have a
moment, the value of which is loa, the resultant is a couple of
moment loa or la since /o is denoted by / when the axis of rotation
passes through the mass-center of the body. The sense of the
resultant couple is, of course, the same as that of a, the angular
acceleration of the body. Further, since the resultant of the
external forces that act on the body is identical with that of the
effective forces for the body, the body may be considered to be in
equilibrium if an additional couple having a magnitude equal to
la and a sense opposite to that of a is assumed to act on the body
with the external forces. This couple is often called the reversed
effective couple or the inertia couple for the body.
If the body rotates about an axis which does not pass through
its mass-center, as indicated in Fig. 350, then f in the above
expressions is not zero. The resultant of the effective forces for
the rotating body, then, is a single force since Mm and Mrco^
express the components of the resultant force, the moment of the
resultant force (or of its two components) being expressed by loa.
The action hne of each of the two components may now be located.
The action Hne of the normal component, Mfcc^, of the resultant
force Ues along the Une connecting the center of rotation and the
mass-center of the body, that is, along the iV-axis as shown in
Fig. 350. This fact may be proved as follows: Mrcc^ was obtained
by resolving the components mai and man, of the effective force
for each particle, parallel to the A^-axis and determining the
algebraic sum of these A^'-components (Art. 146) . And, it was shown
328
FORCE, MASS, AND ACCELERATION
that the man components, only, have an influence in determin-
ing the magnitude Mrco^ since the sum of the mat components
was found to be equal to zero. Now the man component of the
effective force for each particle passes through the center of
rotation. Therefore, the sum, Mfco^, of their components parallel
to the iV-axis must lie along the iV-axis.
The action hne of the tangential component, Mra, of the result-
ant of the effective forces may be found from the principle of
moments; namely, the moment of the resultant of the effective
forces about the axis of rotation equals the sum of the moments of
the effective forces about the same axis. The sum of the moments
Fig. 350.
Fig. 351.
of the effective forces, as already shown, is hoc. Further, the
moment of the resultant of the effective forces is the moment of its
tangential component, Mra, only, since the normal component,
Mr 03^, passes through the center of rotation. Hence, the principle
of moments is expressed by the equation,
Mra • q = loa,
in which q is the moment arm of Mra with respect to the center of
rotation 0, as shown in Fig. 350. And, since h = Mko^, in which ko
is the radius of gyration of the body with respect to the axis of
rotation, we may write,
Mra'q = Mko^a
whence.
SECOND METHOD OF ANALYSIS 329
Therefore, the action lines of the two components, Mra and
Mroy^, of the resultant of the effective forces intersect in a point on
the iV-axis at a distance —^ from the center of rotation, as shown
r '
in Fig. 350. And, since the resultant of the external forces is
identical with the resultant of the effective forces, the body may
be considered to be in equilibrium if the two forces Mfa and Mfcc^,
having the proper action lines as determined above and shown in
Fig. 350, but reversed in sense, are assumed to act on the body
with the external forces. The component Mrco^ of the inertia
force is sometimes called the centrifugal force.
However, instead of determining the resultant of the effective
forces as a single force (or as components of a single force) as was
done above, it is usually more convenient to find the resultant of
the effective forces as an equivalent force and couple; the force
having components equal to Mfa and MrccP parallel, respectively,
to the T- and A^-axes, as before, but with action lines passing
through the mass-center of the body. The fact that the compo-
nents of this new force are the same in magnitude and sense as those
of the single resultant force, follows from Art. 18, in which the
resolution of a force into a force and a couple is discussed. The
fact that the moment of the couple, C, is equal to la may be proved
as follows: The moment of the effective forces about the axis of
rotation is loa as already proved. And, if the resultant of the
effective forces is a force having components Mfa and MfcJ^
acting through the mass-center, and a couple the moment of
which is represented by C, as stated above, then, the principle of
moments is expressed by the equation,
Mfa'f-\-C = Ioa.
But,
/o = /+Mr2 (Art. 102).
Hence,
C = Ia.
Therefore, a rigid body (Fig. 351) rotating about a fixed axis
under the action of an unbalanced force system (P, W, and the
reaction of the axis at 0) may be considered to be in equilibrium
if an additional force and couple are assumed to act on the body
with the external forces. The additional inertia force must act
through the mass-center of the body; it must have components
330
FORCE, MASS, AND ACCELERATION
equal to Mra and Mrco^, the directions of which are parallel
respectively to the T- and iV-axes; and the senses of the com-
ponents must be opposite to those of the corresponding components
of acceleration of the mass-center, as shown in Fig. 351. The
additional inertia couple must have a moment equal to la and
must have a sense of rotation opposite to that of the angular accel-
eration, a, of the body, as shown in Fig. 351. The forces of the
couple, of course, may be assumed to act anywhere in the plane of
motion of the body, provided that the moment of the couple
remains constant (Art. 27).
ILLUSTRATIVE PROBLEMS
375. A horizontal bar B (Fig. 352, a) rotates with a constant angular
velocity of 45 r.p.m. about a vertical axis YY. A slender rod C, of constant
cross-section, having a length of
12 in. and a weight of 16 lb. is
attached to the rotating bar by
means of a smooth pin at E, and
is held in a vertical position by a
weightless cord D. Find the
tension in D and the magnitude
of the reaction of the pin at E on
the rod C.
Solution. — A free-body dia-
gram of the rod C is shown in
Fig. 352, h. The rod has a mo-
tion of rotation about the vertical
axis YY under the influence of
three forces W, D, and the pin
pressure at E (the components
of the pin pressure being denoted
^^^- ^^2. by Ex and Ey).
If the reversed resultant of the effective forces (inertia force) for the rod is
assumed to act on the rod with W, D, Ex, and Ey, the rod may be considered
to be in equilibrium. The resultant of the effective forces is Mrco^, since
a = 0 and hence Mra = 0.
^'""=32:2X2X(-6o-j
= 22.0 lb.
= 22 lb.
161b.
The action line of Mrco^ passes through the mass-center of the rod. Thus,
the forces W, D, Ex, Ey, and Afroj^, as shown in the free-body diagram, would
hold the rod in equilibrium. Using the equations of equilibrium, we have:
SECOND METHOD OF ANALYSIS
331
i:Fx =22-£;x-Z)cos45° = 0
SFy =^y+Dcos45''-16 = 0
2Ms = 22X6-DX9 cos 45° = 0
The solution of these equations gives the following results:
D =20.7 lb.
Ea
7.33 1b., Ey = \.ZZ\h., ^ = 7.45 lb.
376. A bar B, 2 in. by 6 in. by 4 ft., is attached to a cylindrical disc. A,
by means of two steel straps, D and E, as shown in Fig. 353(a). The cylin-
drical disc is keyed to a shaft which causes the disc and the bar to rotate in a
vertical plane. When the bar is in a horizontal position as shown, the angular
velocity, w, of the shaft is 30 r.p.m. and it is decreasing, at the instant, at the
rate of 120 r.p.m. per sec. The weight of the bar is 96 lb. The diameter of
the disc is 18 in. and that of the shaft is 4 in. The straps, D and E, are attached
to the bar by means of smooth pins. Find (a) the torque transmitted from
thie shaft to the bar B, (b) the
sum of the pin pressures in the x"~^<^ h< 2- s4-< z!-
horizontal direction, and (c) the
sum of the pin pressures in
the vertical direction.
Solution. — The resultant of
the effective forces may be con-
sidered to be a force having the
components Mroj^ and Mra: which
act through the mass-center, and
a couple, the moment of which
is 7a. If the effective forces be
reversed as shown in Fig. 353(6)
and assumed to act on the body
with the external forces, the
body may be considered to be
in equilibrium. Hence, the pins must exert a horizontal pressure H{\H for
each pin) to balance the only horizontal force Mr(jo\'ZFx = Q). They must
also exert a vertical force F (^F for each pin) to balance Mra and TF(2Fy = 0).
Finally, the pins must exert a couple C whose moment about any point, G
say, balances the moments of the other forces about the point (2Mg = 0).
The values of the reversed effective forces as shown in Fig. 353(6) are as
follows:
96
32.2'
Fig. 353.
^"^ = 3f2X2-^^x(~6^T = '°-'"''
,,_ 96 ^^ »_^120X27r ,^_,,
""^32^ — 60~ =103 lb.,
7a = j|MZ2a= ix3p2X42X47r = 50.01b.-ft.
332
FORCE, MASS, AND ACCELERATION
Thus, the force system acting on the bar is a non-concurrent system in a
plane, the equations of equilibrium for which are,
i:F^ = 0, 77-80.8 = 0,
XFy = 0, 103-7-96 = 0,
XMg = 0, -C+50.0+2F = 0.
Solving these equations we obtain,
i/ = 80.81b.,
7 = 7.0 lb.,
C = 64 1b.-ft.
The magnitude of each of the forces that form the couple, C, depends on
the distance between the pins. If this distance is 3 in., then each force is 256 lb.
PROBLEMS
377. Solve Prob. 376 by introducing a single force as the reversed resultant
of the effective forces instead of introducing a reversed force and couple.
378. A small body is placed on a rough horizontal disc which rotates about
a vertical axis. If the distance of the body from the axis is 9 in. and the
coefficient of friction between the body and disc is f, find (1) the greatest
ang liar velocity and (2) the greatest angular acceleration the disc can have
without causing the body to slide. Ans. u)=51 r.p.m ; a =28.5 rad./sec.^
379. A flywheel used on a punching machine is 8 ft. in diameter and has a
rim which weighs 1 ton. Each operation of punching a hole causes the speed
of the flywheel to decrease uniformly
from 100 r.p.m. to 80 r.p.m. The fly-
wheel has 6 spokes, each 3.5 ft. long. If
the time of punching a hole is 0.5 sec,
what moment is transmitted from the
rim to the hub by each spoke. Assume
that the thickness of the rim is small in
comparison with the radius of the fly-
wheel and neglect the weight of the hub
and spokes. Ans. 607 Ib.-ft.
380. A uniform bar AB and a mem-
ber C, to which AB is pinned at A (Fig.
354), rotate, at a speed of 40 r.p.m.
about a vertical axis. The weight
of the bar is 40 lb. and its length is 4 ft.
If the bar is connected to the axis by a
Fig. 354. spring S (assumed horizontal), as shown
CENTER OF PERCUSSION
333
in the figure, find the pull of the spring and the horizontal and vertical
components of the pin pressure at A.
381. In Fig. 355, A represents a
frame which revolves about a vertical
axis FF at a constant angular velocity
CO =40 r.p.m. A bar, B, is attached to
the frame at E by means of a smooth
pin. At the end of 5 a spherical ball, C,
is fastened. B weighs 20 lb. and is 16
in. long. C weighs 8 lb. and is 4 in. in
diameter. Find the reaction of the pin at
E and of the frame at F, on the bar.
-80^
'^
Ifl
^
148. Center of Percussion. —
The point P (Fig. 356) on the
A^-axis, through which the result- pn. 355
ant of the effective forces for a
rotating rigid body acts, is called the center of percussion of the
body with respect to the given axis of rotation. Hence, the center
of percussion is a point on a line joining the center of rotation and
the mass-center, at a distance q from the center of rotation,
such that,
A-o2
9 = ^>
in which A;o is the radius of gyration of the body about the axis
334 FORCE, MASS, AND ACCELERATION
of rotation and r is the distance from the axis of rotation to the
mass-center of the body.
The physical significance of the center of percussion is suggested
in the following illustration. Let a bar (Fig, 356) of weight W
be free to rotate about a horizontal axis when a horizontal
force, F, is suddenly applied to it. If the force, F, is applied
above the center of percussion, as shown in Fig. 356(a), the hori-
zontal reaction, R2, of the axis of rotation acts towards the left
and becomes larger as the force F is applied closer to the axis of
rotation. If the bar is struck below the center of percussion, the
reaction R2 acts towards the right, as shown in Fig. 356(6).
And if the bar is struck so that the center of percussion is on the
action line of the force, as in Fig. 356(c), the horizontal reaction at
0 is zero, since the action hne at F is collinear with the action line
of the tangential component, Mra, of the resultant of the effective
forces. It will be noted that the resultant of F and R2, in each
case, is collinear with Mfa, since the component of the resultant of
the external forces in any direction is identical with the component
of the resultant of the effective forces in the same direction, that is,
if Mra were reversed and applied to the body as an external force,
it would hold F and R2 in equilibrium.
An excellent illustration of the effect of varying the position of
the force F as above discussed is found in batting a baseball. If
the ball strikes the bat at the center of percussion (about three-
fourths the length of the bat from the end, assuming the axis of
rotation at the hands) no reaction perpendicular to the bat is
experienced by the batter. If, however, the ball strikes the bat
near the end or near the hands, the batter experiences a painful
stinging of the hands as a result of the reaction perpendicular
to the bat.
ILLUSTRATIVE PROBLEM
382. A slender rod AB is caused to rotate about its lower end A by the
crank CD, the length of which is 1 ft., and the link DE (Fig. 357a). The
angular velocity, co, of the crank is very small at the given instant but its
rate of change, a, is 20 rad. per sec.^ The weight of the rod AB is 64.4 lb.
and its length, I, is 5 ft. A small body F which weighs 16.1 lb. is attached
to the rod at a distance of 2 ft. from A. Body F may be considered to be a
particle. What is the distance, d, from A at which the link DE must be
attached if no horizontal reaction occurs at ^ ?
Solution. — The resultant of the effective forces for the rod is Mrai (where
ai is the angular acceleration of the rod) since wi is very small and hence
CENTER OF PERCUSSION
335
MFm^ is negligible. The tangential acceleration of E, as a point on AB, is
dai and the tangential acceleration of D, as a point CD, is CD a. Since the
accelerations of D and E along ^D are equal, we have (Art. 121),
Hence,
(a)
dai=a, since CD = 1 ft.
« 20
B
'>r
«f^
V/o =64.4 lb.
1^ -.
U w 1'
t
t Wi = 16.1
-fib
a
lb.
T
(ft)
Fig. 357.
Therefore, the resultant of the effective forces for the rod is.
The distance of the action line of Mrai from A is.
/c/
E
^i
:i = 3.33ft.
The effective force for particle F is,
16.1
20 20
100
and its action line passes through the particle F. If these two forces, —7- lb.
a
20
and -r lb., are reversed and assumed to act on the rod with the forces P, Ry,
Wi, and TF2, as shown in Fig. 357(6), the rod may be assumed to be in equilib-
336 FORCE, MASS, AND ACCELERATION
rium. The value of d may be determined by using only two of the equilib-
rium equations. Thus,
ad'
SM4=Pd-^X3.33-?5x2=0.
d d
Solving the equations, we have,
P= 38.6 lb.,
d = 3.11 ft.
PROBLEMS
383. A uniform slender rod, 6 ft. long and weighing 20 lb., is suspended
from a horizontal axis at one end and is acted on by a horizontal force of 20
lb. at its mid-point. Determine (a) the resulting angular acceleration, (6) the
resulting linear acceleration of the mass-center, (c) the horizontal reaction of
the axis on the rod, and {d) the distance from the axis at which the force must
be applied so as to cause no horizontal reaction.
Ans. (a) 8.05 rad. per sec.^ (c) 5 lb.
(b) 24.15 ft. per sec.^ (d) 4 ft.
384. A door of constant cross-section is 3 ft. wide and weighs 32.2 lb. per ft.
of width. It swings on its hinges so that its outer edge has a speed of 8 ft.
per sec. (a) Find the force applied perpendicularly to the door at the outer
edge to .bring it to rest in a distance of 1 ft. (6) What is the horizontal reaction
of the hinges perpendicular to the door while the force is acting? (c) How far
from the hinge line must the force be applied in order that the hinge reaction
shall have no horizontal component perpendicular to the door?
Plane Motion
149. Kinetics of Plane Motion of a Rigid Body. — As shown in
Art. 134, a plane motion of a rigid body may be considered, at any
instant, as a combination of a rotation about an axis through any
point, 0, in the plane of motion of the body, and a translation of
the body which gives to each particle the same velocity and
acceleration that the point 0 has at the instant. The motion of
any particle of the body, therefore, may be resolved into two
component motions, (1) a rotation about 0 and (2) a motion iden-
tical with that of 0. Hence, the acceleration of any particle
has a normal component, an = r(J^, and a tangential component,
Ot = ra, due to the rotation of the body about 0, and also an acceler-
tion, oo, the same as that of 0, due to the translation of the body.
KINETICS OF PLANE MOTION OF RIGID BODY 337
Thus, in Fig. 358, let the diagram represent a body which has
plane motion. The body is assumed to be symmetrical with
respect to its plane of motion and hence may be considered to be
concentrated in the plane of motion (see footnote under Art. 144).
Let the angular velocity and angular acceleration of the body, at
the given instant, be cc and a, respectively (co and a are the same
with respect to all axes perpendicular to the plane of motion and
all particles have the same co and « at a given instant, Art. 134).
Let the motion be resolved into a rotation about an axis through
any point 0, giving to the body, of course, its actual angular
w
Fig. 359.
velocity and acceleration, and a translation which gives to each
particle a motion the same as that of 0. Let oo denote the acceler-
ation of the point 0. Then, the acceleration of any particle
at a distance r from 0 (Fig. 358) has three components; rco^, ra, and
Oo, the directions of which are shown for several particles in the
figure.
After the components of the acceleration of any particle of the
body have been found, the components of the resultant force (effect-
ive force) required to produce the component accelerations may be
found from Newton's second law. Thus, as shown in Fig. 359,
the components of the effective force for any particle of mass m
are mat = mra in the direction of at, man = mroy^ in the direction of
an, and moo in the direction of oq. For convenience, mao will be
resolved into its x- and ^/-components, m(ao)x and m{ao)y, as shown
in Fig. 359.
338 FORCE, MASS, AND ACCELERATION
Resultant of Effective Forces. — The resultant of the effective
forces for the whole body may now be found. The effective forces
form a non-concurrent system in the plane of motion. The result-
ant of such a system of forces is either a force or a couple (Art. 28).
It will be shown, however, that when the resultant is a force it
may be resolved, for convenience, into another force which acts
through the mass-center of the body and a couple.
The resultant of a non-concurrent force system in a plane may
be determined completely from the algebraic sums of the o^-com-
ponents of the forces, of the y-components of the forces, and of
the moments of the forces about an axis in the body through 0
(Art. 30). Thus, referring to Fig. 359, and considering the
effective forces for the whole body we may write:
The algebraic sum of the a:-coniponents of the effective forces
= 2m(ao)z+2wra sin d — llmoy^r cos 6
= (ao)z2m+Q;2m7/— co^Smo;
= M{ao):c-{-Mya-MxoP.
The algebraic ^sum of the ^/-components of the effective forces
= 2m(ao)i/ — 2mm cos 6— 'EmoPr sin d
= (ao)vSm— aSmx— co^Zm?/
The algebraic sum of the moments of the effective forces about 0
= Smra • r-\-Xm(ao)xy — 'Lm{ao)yX
=a'Emr^-\-{ao)x^my—{ao)yi:mx
= Ioa-{-My{a^):c- Mx{ao)v.
But the algebraic sum of the a:-components of the effective forces is
equal to the o^-component of the resultant of the effective forces,
and similarly for the ^/-components. Further, the algebraic sum of
the moments of the effective forces with respect to 0 is equal to
the moment of the resultant of the effective forces with respect to 0.
Hence, summarizing.
The a:-component of the resultant of the effective forces
= M (do) x+Mya-Mxoi^.
KINETICS OF PLANE MOTION OF RIGID. BODY 339
The i/-component of the resultant of the effective forces
= ilf (oo) y — Mxa — MyJ^.
The moment of the resultant of the effective forces about 0
= Ioa+My{(U)):c-Mx{ao)v.
Relation between External and Effective Forces. — Since the effect-
ive force for a particle of a body is the resultant of all the forces
acting on the particle, it is the resultant of both the internal and
the external forces which act on the particle. Therefore, the
component, in any direction, of the resultant of the effective forces
for the whole body is the algebraic sum of the components, in the
given direction, of all the internal forces and all the external forces
which act on the particles. Likewise, the moment of the resultant
of the effective forces about any axis perpendicular to the plane
of motion is the algebraic sum of the moments of all the internal
forces and all the external forces which act on the particles.
Hence, if F denotes a force and T denotes the moment (torque)
of a force, the above expressions are connected with the forces
acting on the particles according to the following equations:
i^Fx) Extcrnal+(2i^rr) Internal = M(ai))x+Mya- Mxoi^
i^Fy) Bnernal+C^Fy) internal = M (oq) y - Mxa - My 03^
(2To) External+CSTo) Internal = /o« + M^(ao)x-Afx(ao)^.
But, since the internal forces occur in pairs of equal, opposite,
and coUinear forces (Newton's third law), the algebraic sum of
their components in any direction, and the algebraic sum of
their moments about any axis perpendicular to the plane of motion,
are zero (that is, the resultant of the external forces alone
and the resultant of the effective forces for the body are identical).
Hence, in the above equations, the terms which involve the internal
forces may be omitted and therefore the equations expressing the
relation between (1) the external forces acting on the body, (2) the
kinetic properties of the body, and (3) the change in motion of
the body are :
l^Fx = Miao).+Mya-Mx<o^
:LFy = M{ao)y-Mxa-My(o^
STo = Ioa-i-My{ao)x-Mx{ao)y
(1)
340
IIORCE, MASS, AND ACCELERATION
The equations of motion, however, may be written in a simpler
form. As already noted, the center, 0, about which the assumed
rotation takes place and about which the moments of the forces
are taken may be any point in the plane of motion of the body.
Thus, if the mass-center is selected for the center about which
moments are taken, that is, if 0 coincides with G (Fig. 359) then,
in the above equations, a: and y are zero; ao becomes a; lo becomes
J; and 2 To becomes XT. Hence the right-hand member of each
of the above equations reduces to one term. Thus the equations
of motion for a rigid body having plane motion may be written :
Si^x =
= Max
^Fy =
= May
sr=
= la
(2)
ILLUSTRATIVE PROBLEMS
385. A homogeneous cylinder which is 3 ft. in diameter and which weighs
805 lb. rolls down an inclined plane which makes an angle of 30° with the
horizontal (Fig. 360). The
mass-center of the cylinder has
an initial velocity yo = 50 ft.
per sec. The plane is rough so
that the cylinder rolls without
slipping. Find (1) the acceler-
ation of the mass-center, (2)
the magnitude of the friction
force, and (3) the velocity, v,
of the mass-center at the end of
W -805 lb. 10 sec.
Solution. — The cylinder has
plane motion under the action
Fig. 360. of three forces, F, N, and W,
as shown in Fig. 360. Let the
X- and y-axes be chosen as shown in the figure. The equations of motion are,
i:Fx = Maj, (1)
XFu=May, (2)
i:T = Toc (3)
805sin30°-i^ = ^az (4)
From (1),
KINETICS OF PLANE MOTION OF RIGID BODY 341
From (2),
From (3),
■805 cos30°+A^ = 0, since ay = 0.
3 1 «nf; /^\ 2
1 805 /3
2 2^^32.2
©-
(5)
(6)
There are four unknown quantities involved in these three equations and
hence another equation is needed. Frotn the kinematics of the problem
(Art. 121), we obtain the equation,
ax=a=ra=^a (7)
By substituting the value of a from (7) in (6) and solving for F, we obtain the
equation,
Substituting this value of F in (4), we obtain,
aa; =a = 10.73 ft./sec.2
Hence,
/f = |^Xl0.73 = 134.1 lb.
Since the mass-center moves with uniformly accelerated rectilinear motion,
we may use the equation,
v=Vo-\-at,
Hence,
iy = 50+10.73X10
= 157.3 ft./sec.
386. The connecting rod of a steam (^.ngine has a length Z=6 ft. and a
weight TF = 220 lb. (Fig. 361). The length of the crank is r = l ft. and the
Fig. 361.
engine runs at a speed of coi=300 r.p.m. The pressure P on the cross-
head, when 0 = 30°, is 18,000 lb. Assume that the cross-section of the con-
necting rod is constant. Find the tangential crankpin pressure, Q, the com-
342 FORCE, MASS, AND ACCELERATION
pression, S, in the crank arm, and the guide pressure, A^, which is assumed to
act vertically (friction neglected).
Solution. — The connecting rod has plane motion. Either equations (1)
or equations (2) of Art. 149 may be used. Let equations (1) be chosen and let
the crosshead end of the connecting rod be selected as the origin. Further,
let the X- and y-axes be chosen horizontal and vertical respectively. The
equations of motion then are,
XFx = M(aA)x+Mya-Mxo}^, (1)
XFv^M(aA)v-Mxoc-Myo:^, (2)
7:TA=lAa+My{aA)x-Mx{aA)u, (3)
in which,
220
{aA)v=0,{aA)x=aA, M = |~=6.84 slugs, Ia=^MI'=S2 slug-ft.^
Further,
I sin 4>=r sin 6 or sin «/> = | sin 30° = 1-2.
Hence,
<^=4° 47'.
Therefore,
sin 0 = .0833, x = 3 cos </> = 2.99 ft.,
cos (f> = .9965, ^ = 3 sin (/> = 0.25 ft.
ttA, CO, and a may be found by use of the semi-graphical method discussed in
Scales:- 9 *
i"-2ft. •! I z^-
1"— 30 -'t./sec. 1 //
l"-400 ft./sec.2 ^. /^-^
1 /^^
C3t<
A ^ ^-^^ ^"^
aA— 936 (scaled) ^^\ T ^<
b/ (27.4)-
■^k-'-t-'T-^
Fig. 362.
Prob. (334). Thus, in Fig. 362, the following two equations, (4) and (5),
are solved,
«'fi=^^4>'^4' (4)
A
KINETICS OF PLANE MOTION OF RIGID BODY 343
n
(w / V , (27.4)2 ,. ^ /300X27rV ,_,
i;^ ^A^~^5> (^a\ ~~ (^b\ ^^^ ^A ^^® found, by scaling off their
B A \b) t \^) I
lengths, to have the following values:
y^ = 18ft./sec.,v^=27.4ft./sec., (as) =492ft./sec.2, a^ =936 ft. /sec.''
A ^ A^'
But.
Vjj = o:l, and {^b\ =al.
A ^^)'
Hence,
<o=— r^ =4.56 rad. /sec. and, q; = -^ =82 rad./sec.^.
o D
From (1),
18,000-*S cos 30°-Q cos 60° = 6.84 (936+0.25X82-2.99X4^').
From (2),
N+S sin 30°-Q sin 60°-220=6.84 (0-2.99X82 -0.25 X4L56')-
From (3),
-3.42 >Sf+4.93 Q+2.99X220 = 82X82+6.84X0.25X936.
Solving for the three unknown forces, we obtain the following values:
Q = 7950 1b., AS:=9150 1b., iV = 795 1b.
PROBLEMS
387. A homogeneous solid sphere rolls without slipping down a reugh
plane which is inclined at an angle 9 with the horizontal. Show that the
acceleration of the center of the sphere is ^g sin 6, and that the ratio of the
friction to the normal pressure must be not less than f tan 6 to prevent the
sphere from slipping.
344
FORCE, MASS, AND ACCELERATION
388. A solid sphere having a radius of 8 in. and a weight of 161 lb. is made
to roll up a rough inclined plane (Fig. 363) by means of a flexible cord, one
end of which is attached to an axis through the center of the sphere. The
cord passes over a smooth peg and has attached to its other end a suspended
body B which weighs 100 lb. Find the acceleration of the body B and the
tension in the cord. Ans. a = 1.93 ft. /sec. ^
Fig. 363.
Fig. 364.
389. A homogeneous cylinder 1 ft. in diameter has a flexible cord wrapped
around its central plane. One end of the cord is attached to a fixed plane as
shown in Fig. 364. The cord is taut when the cylinder is allowed to fall.
Find (a) the acceleration of the mass-center, (6) the angular acceleration of the
cyhnder, and (c) the distance traveled by the mass-center in 10 sec.
390. Solve Prob. 386 by use of equations (2) of Art. 149.
391.
A uniform rod AB (Fig. 365) moves with its ends B and A in contact
with smooth planes which are vertical
and horizontal respectively. A variable
horizontal force, F, is applied at the end
A. The weight of the rod is 16.1 lb.
and its length is 8 ft. If the value of F
for the position of the rod shown (0 = 60°)
is such that the angular acceleration, a,
of the rod is 3 rad./sec.^, and the
angular velocity, w, is 2 rad./sec, what
are the values of F, Na, and Nb?
Ans. iVA = 6.171b.; F=-.03Ib.;
Nb = 1.22 Ih.
Fig. 365.
its central plane and a force,
392. Two solid cylindrical discs are
keyed to an axle as shown in Fig. 366.
A string is wrapped around the axle in
P, is exerted by the string in a directicji
SECOND METHOD OF ANALYSIS _ 345
parallel to the plane on which the discs roll and tangent to the under surface
of the axle. Each disc weighs 20 lb. and is 2 ft. in diameter. The axle is
6 in. in diameter and weighs 40 lb. The magnitude of the force P is 8 lb.
W , 11 the discs and axle roll forward or backward? Find the acceleration of the
central axis of the discs and axle. Arts, a = 1.91 ft. /sec. ^
393. If the string in the pre-
ceding problem is wrapped around
the axle in the opposite direction
so that the force P is tangent to the
top of the axle, what is the acceler-
ation of the central axis of the
discs and axle?
394. A wheel has an eccentric
weight attached near its rim so that
its mass-center is not the geometric
center of the wheel. If the wheel Fig. 366.
rolls on a horizontal track, will
the wheel leave the track when its velocity attains some particilar
value?
150. Second Method of Analysis. Inertia Forces.— In some
problems which deal with the plane motion of a rigid body under
the action of an unbalanced force system, it is convenient to assume
that the reversed resultant of the effective forces (inertia force for
the body) acts on the body with the external forces, thereby forming
a force system that is in equilibrium and thus reducing the kinetics
problem to an equivalent statics problem. In order to use this
method of analysis, the location of the action line of the resultant
of the effective forces must be known, as well as the magnitude
and sense of the resultant force.
Equations (1) of the preceding article state that the resultant
of the effective forces for the bod}^ (and hence also of the external
forces which act on the body) is a force having x- and ^/-components
equal to M{ao)x+Mya — Mxcxy^ and M{ao)y — Mxa — Myo?, respect-
ively, as shown in Fig. 367(a), and that the moment of the resultant
force with respect to the axis through the origin 0 is equal to
Ioa-\-My{ao) X — Mx{ao)y. Equations (2) state that the resultant
of the effective forces for the same body is a force having x- and y-
components equal to Max and May, respectively, as shown in Fig.
367(6), and that the moment of the resultant force with respect to
an axis through the mass-center of the body is la. The resultant
of the effective forces for any given body as obtained from each set
346
FORCE, MASS, AND ACCELERATION
of equations is, of course, the same force ^ and its components may
intersect at any point on its action line such as points A and A' in
Fig. 367(a) and (b).
The action hne of the resultant force as located by its moment
arm ^o in Fig. 367(a) or q in Fig. 367(6) may be found from the
principle of moments by equating the moment of the resultant of
Fig. 367.
the effective forces to the sum of the moments of the effective
forces. The sum of the moments of the effective forces about the
mass-center, as already found, is la. And, the moment of the
resultant force having the components Max and May is Maq
(Fig. 3676). Thus the principle of moments is expressed by the
equation,
Ma'q = Ia. But I = Ml?.
Therefore,
^ Thus we may write,
M{ao)x+Mya—Mxco^ = Max,
M(ao)y—Mxa — Myu,^ = May.
Or, canceling the M's, we have the equations dealing with accelerations as
follows,
(cio)x+ya—xo}^ = ax,
iO'o)v—xa — yo}^ = ay,
which state that the acceleration of the mass-center is equal to the relative
acceleration of the mass-center with respect to any axis, O, plus the accelera-
tion of O; a proposition which is discussed in Art. 128.
SECOND METHOD OF ANALYSIS
347
The value of qo (Fig. 367a) may be found in a similar manner.
However, it is more convenient, in general, to resolve the single
resultant force Ma, having an action line which is located at the
distance q = -^- from the mass-center, into another force equal
and parallel to Ma, but with its line of action passing through the
mass-center, and a couple, C, having a moment equal to Ma-q
(Art. 18) as shown in Fig. 368(a) . But, as shown above. Ma -q^Ia
and hence the moment of the couple is la.
Fig. 368.
Therefore, the resultant of the effective forces for a rigid body
having plane motion may be considered to be a force Ma having
an action line which passes through the mass-center of the body
in the direction of the acceleration, a, of the mass-center, and a
couple having a moment equal to la, the sense of the couple being
the same as that of the angular acceleration, a, of the body. The
above facts together with D'Alembert's principle lead to a state-
ment or proposition which is of great importance in the method
of reducing the kinetics problem of a rigid body having plane
motion to an equivalent problem in equihbrium by introducing
the inertia force (or forces) for the body. The proposition may
be stated as follows:
If a rigid body is acted upon by an unbalanced force system
which gives to the body a plane motion, the body may be con-
sidered to be in equilibrium if an additional force and couple is
assumed to act on the body with the external forces; the addi-
tional force has a magnitude equal to Ma, its line of action
348
FORCE, MASS. AND ACCELERATION
passes through the mass-center of the body, and its direction
is opposite to that of a; the couple has a moment equal to la
and its direction of rotation is opposite to that of a.
The forces of the couple are not necessarily equal to Ma or
parallel to a as shown in Fig. 368(a), that is, the couple may be
rotated in the plane of motion and the magnitude of the forces
of the couple may be changed provided the moment of the couple
is always equal to la (Art. 27). Thus, if a force having a magni-
tude equal to Ma, and a couple having a moment equal to Ma • q
(or la), act on the body with the external forces (P and W), as
shown in Fig. 368(6), the force system (and the body) would be in
equilibrium.
ILLUSTRATIVE PROBLEMS
395. At what height should the cushion on a bilHard table be placed so
that the billiard ball in rebounding from the cushion starts off without causing
any friction on the table top?
Solution. — The billiard ball has plane motion and while in contact with the
cushion it is acted upon by three forces ; the cushion pressure P, the weight W,
and the table pressure N, as shown in Fig. 369.
If there is no horizontal friction force at A, then P is the resultant of
the external forces which act on the ball and it must be collinear with the
resultant of the effective forces. The re-
sultant of the effective forces is a force of
magnitude Ma, the distance of the action
line from the mass-center being given
by the equation,
Q =
And since a =ra, we have
Hence,
_lt^_ir^
5
396. Solve Prob. 386 by introducing the reversed resultant of the effective
forces (inertia force for the rod).
Solution. — The resultant of the effective forces is a force of magnitude M'a
acting through the mass-center of the rod in the direction of a, and a couple
of moment la, as shown in Fig. 370(a). If G denotes the mass-center, then
SECOND METHOD OF ANALYSIS
349
a may be written ao for convenience, og niay be found from the graphic
equation,
1/ 2
2 t-
in which i-^ = *y^ = 13.7 and a = 82. (See Prob. 386.) Hence,
B B
«(y — 3— -B>3X82-B>986
= 62.6 +> 246 -f> 986
= 930ft./sec.2,
as scaled from the acceleration polygon of Fig. 370(a).
Scales^-
a"=2fft.
1 = 400 ft. /sec?
Therefore the resultant of the effective forces is a force,
M
and a couple,
220
Ma =11^X930 = 6360 lb.,
But,
1 1 220
Ioc=j^Ml^a = j^X^^X6'X82 = lQ8.1 Ib.-ft.
Ia = Ma'q or 168.1 =6360g. /. 2= 0.264 ft.
350 FORCE, MASS, AND ACCELERATION
Hence q = .264 ft., and the inertia force for the rod is a single force of magnitude
6360 lb. as shown in Fig. 370(6). If the inertia force is assumed to act on the
rod with the external forces, the rod may be considered to be in equilibrium.
The three unknown forces may be found graphically by use of the force and
funicular polygons or, the values of /3 and e may be scaled off and the equations
of equilibrium used. The values of /3 and e are found to be /3 = 15° 35' and
e = 0.77 ft.
The equations of equilibrium for the non-concurrent forces in a plane as
shown in Fig. 370(6) are,
SFx = 0, 18,000-6360 cos /3-/S cos 30°- Q cos 60° = 0, . . (1)
SFi,=0, iV - 220 +6360 sin |8 +5 sin 30° -Q sin 60° =0, . . (2)
SMfi=0, 18,000 Z sin 0-M cos 0+220X2.99-6360X0.77 = 0, (3)
in which sin (f> and cos <f> have the same values as were used in Prob. 386.
Thus,
sin (^ = .0833, cos 0 = .9965, sin /3=sin 15° 35' = .268, cos )S = .963, 1 = 6 ft.
Substituting the known values in the above equations, we obtain the following
equations:
.5Q+.866>S- 11,875=0,
.866Q-.5/S -2282=0,
-5.98A^ +4758=0,
the solution of which gives the following values for the three unknown forces :
Q = 79701b.; ^ = 9100 lb.; iV = 7971b.
PROBLEMS
397. A solid homogeneous disc 3 ft. in diameter rolls on a straight hori-
zontal track (Fig. 371). A small body, B, is attached to the disc at a distance
d = 0.75 ft. from the center of the disc.
When in the position shown in Fig. 371
the angular velocity of the disc is 6 rad.
per sec, and a force P is retarding the
angular velocity at the rate of 4 rad.
per sec. each second. The disc weighs
60 lb. and B weighs 8 lb. Find the
'^y.y////////y/yy^^^^^^^ \T*^J ^""ri ^""^ ^^^ disc and for the body £.
Also find the value of F and of the friction
Fig. 371. force. Ans. P = 7.91 lb.; i^ = 1.96 1b.
398. The resultant, i?, of the forces acting on the connecting rod (Fig.
372) is 320 lb. and its action line is located as shown. If the connecting
LIMITATIONS OF MOMENT EQUATION T = Ia 351
rod is 30 in. long and weighs 83 lb , what is the linear acceleration of the
mass-center of the rod and the angular acceleration of the rod, assuming
the rod to be of constant cross-section?
Fig. 372.
151. Limitations of the Moment Equation 7 = /a.— As dis-
cussed in Art. 134, a rigid body having plane motion may be given
its actual motion, at any instant, by a combination of a rotation
and a translation. And, it was shown that in giving the body the
rotational motion, the axis about which the body is assumed to
rotate may be any axis, 0, in the body, perpendicular to the plane
of motion, the corresponding translation of the body being
defined by the motion which the arbitrarily chosen axis, 0, has at
the instant. It does not follow, however, that the resultant
external force which is required to produce the translation is
the product of the mass of the body and the translatory accelera-
tion, oo, which each particle receives, as is the case for a body for
which the actual (or sole) motion is translation (Art. 145). Nor
does it follow that the moment of the resultant of the external
forces which is required to produce the rotational part of the
motion is the product of the moment of inertia, Jo, of the body
about the arbitrarily chosen axis and the angular acceleration,
a, of the body, as is the case for a body for which the actual (or
sole) motion is rotation (Art. 146).
Equations (1) of Art. 149 show that the resultant force required
to produce the translatory part of a plane motion is the product
of the mass of the body and the translatory acceleration, only
when the axis 0 is chosen through the mass-center of the body,
that is, when x and y in equations (1) are zero. The simple moment
equation, J'o = /o«, also applies to a rigid body having a plane
motion when the arbitrarily chosen axis, 0, passes through the
mass-center of the body, as was also shown in Art. 149.
It also applies, however, for two other positions of the axis 0,
352 FORCE, MASS, AND ACCELERATION
as stated under (3) and (4) below. Thus, in considering a plane
motion of a rigid body under the action of an unbalanced force
system, the moment of the resultant of the external forces about
an axis in the body is equal to the product of the moment of inertia
of the body with respect to the same axis and the angular accelera-
tion of the body only when one (or more) of the following four
conditions are satisfied, the first of which restricts the motion
of the body to a special case of plane motion (rotation) and the
other three restrict the choice of the axis in the body, about which
the rotational part of the motion is assumed to take place :
1. If the body rotates about an axis which is fixed in space
as well as in the body, then ao, in equations (1) of Art. 149, is
zero and hence the moment equation reduces to To = Ioa.
This restriction on the axis limits the motion to pure rotation,
which is treated in Art. 146 and is, of course, a special case of
plane motion.
2. If the arbitrarily chosen axis in the body, about which
moments are taken, passes through the mass-center of the
body, then x and y in equations (1) are zero and the moment
equation reduces to T=I(x as discussed above.
3. If the acceleration of the point 0 (in the plane of motion)
through which the moment axis is taken, is directed toward
the mass-center of the body, then, the quantity lM{ao)xy
— M(ao)vx] in the expression for the moment, in equations (1),
is zero and hence the moment equation reduces to To = Ioa.
The quantity [M{ao)xy—M{ao)yx] is the moment of the force
Moo with respect to the mass-center of the body, and if oo
passes through the mass-center, then Mao also passes through
the mass-center and hence its moment with respect to the
mass-center is zero.
4. If the arbitrarily chosen axis in the body, about which
moments are taken, passes through the instantaneous center
of zero acceleration (not the instantaneous center of zero
velocity. Art. 135) then oo in equations (1) is zero and the
moment equation reduces to To = loa.
The location of the arbitrarily chosen axis as required under
(3)' and (4) above is of considerable importance in the study of
certain problems in Dynamics of Machinery, a detailed discussion
of which is beyond the scope of this book.
PRINCIPLE OF MOTION OF MASS-CENTER 353
PROBLEM
399. Solve Prob. 385 by use of the equations; XFx = Max, ^Fy = May, and
'STA=lAa, in which moments are taken about an axis through A, the point
of contact of the cyUnder with the inchned plane. Does the moment equa-
tion STA=/^a apply for this position of the moment axis because A is the
point about which the body is rotating at the instant (instantaneous center of
zero velocity)? If not, which one of the four special cases stated above applies
when the moment axis passes through A?
152. Principle of the Motion of the Mass-Center. — For
each of the three types of motion of rigid bodies (translation,
rotation, and plane motion) already discussed in this section,
there are three equations of motion. The first two of each of these
sets of equations ((2) of Art. 145; (1) of Art. 146; and (2) of
Art. 149) may be written so as to apply to all three types of motion.
The equations are,
H^Fx^Max and J:Fy = Mav,
in which x and y denote any two axes at right angles to each other.
Thus, for the motion of translation of a rigid body, cix and ciy are
the same as the ax and ay of any other point in the body and hence,
in equations (2) of Art. 145, ax = ax and ay = ay. For the motion of
rotation of a rigid body the T- and A^-axes were chosen and in the
equations (1) of Art 146, fa = at and rco^ = an- For plane motion
of a rigid body the first two of equations (2) of Art. 149 are already
in the form written above. However, these two equations are
not restricted to the motion of rigid bodies having the three types
of motion treated in the preceding articles. They apply to any
mass-system having any type of motion. , They express mathe-
matically a principle which is called the principle of the motion of
the mass-center. The extension of the proof of this principle to
any motion of any mass-system involves the same fundamental
laws and the same methods as were used in the preceding articles
in deriving the equations of motion for a rigid body having the
three special types of motion. The principle of the motion of
the mass-center may be stated in words as follows:
If an unbalanced external force system acts on a body
(whether rigid or not), the resultant of the external force
system, if a force, has a magnitude which is equal to the
product of the mass of the body ^nd the acceleration of th^
354 FORCE, MASS, AND ACCELERATION
mass-center of the body, and the direction of the resultant
force is the same as that of the acceleration of the mass-center.
Likewise, the component, in any direction, of the resultant of
the external forces is equal to the product of the mass of
the body and the component of the acceleration of the
mass-center in the given direction.
The statement of the principle as given above is in the form
which makes it most directly apphcable to the method of analysis
which reduces the kinetics problem to an equivalent statics prob-
lem by introducing inertia forces. The principle may also be
stated in the following form :
The algebraic sum of the components, in any direction,
of the external forces which act on any mass-system is equal
to the mass of the whole body times the component, in the
given direction, of the acceleration of the mass-center of the
mass system.
It will be noted that, according to the principle of the motion
of the mass-center, the mass-system or body may be assumed
to be reduced to a particle having a mass which is equal to the total
mass of the body and which is located at, and moving with, the
mass-center of the body. That is, the magnitude and direction of
the resultant of the actual (non-concurrent) forces which act on
the actual body are the same as the magnitude and direction of the
resultant of the (concurrent) forces which would have to act on
the mass if it were concentrated at, and moving with, the mass-
center of the body.
It is important to note, however, that the action line of the
resultant of the forces acting on a body, does not, in general, pass
through the mass-center of the body whereas it would necessarily
do so if the mass of the body were actually concentrated at the
mass-center. Further, the principle cannot be used to determine
the resultant of the forces acting on a body when the resultant is a •
couple. Nevertheless, the principle simplifies many problems
and is of much importance in the study of kinetics.
PRINCIPLE OF MOTION OF MASS-CENTER _355
ILLUSTRATIVE PROBLEM
400. Two bodies A and B (Fig. 373) are connected by a light slender rod
and revolve in a horizontal plane about an axle fixed in the top of a vertical
post which supports the two bodies.
Body A weighs 4 lb. and B weighs 12 lb.
The bodies rotate with an angular
velocity of 80 r.p.m. What horizontal
force acts on the post tending to bend
the post? The mass of the rod may be
neglected.
Solution. — The mass-center of the
two bodies is found by the principle Fig. 373.
of moments to be at a distance r = 5 in.
from the axis of rotation. The acceleration of the mass-center is,
.. .. ,-. /80X27r\2 5 ^„^ ,. ,
a=an = co^r= I j Xy2=29.2 ft./g
directed toward the center of rotation. The resultant force acting on the
two bodies is,
Ma„ = ^^-X29.2 = U.5 lb.
This force is exerted on the two bodies by the post, and the two bodies exert an
equal and opposite force on the post. (Check the answer by finding the force
acting on each body and subtracting one force from the other.)
PROBLEMS
401. A solid wooden disc 10 ft. in diameter rotates in a horizontal plane
about its geometric axis. Two small bodies each weighing 50 lb. are attached
to the disc at a radius of 4 ft. from the axis of rotation so that the radii make
an angle of 90°. If the disc rotates at 40 r.p.m. what is the resultant horizontal
pull on the axis? Solve by two methods, Ans. 154 lb.
402. A flat-topped boat having a weight of 300 lb. and a length of 12 ft. is
resting in still water. A man weighing 150 lb. stands at one end of the boat.
The man starts to run with a speed increasing at the rate of 10 ft. per sec.
each second. When he reaches the other end of the boat he jumps. Assuming
that the water is a perfect (frictionless) fluid what is the acceleration of the
mass-center of the boat and man (considered as one body) before the man
starts to run? While he is running on the boat? After he jumps from the
boat but before he strikes the water? What is the acceleration of the boat
while the man is running on it?
403. Two bodies A and B (Fig. 374), each weighing 8 lb., are connected by a
string. Body A is placed on a smooth table and the other body B is sus-
356
FORCE, MASS. AND ACCELERATION
pended over the edge of the table. The two bodies start from rest and move
two seconds before body A reaches the table edge. What is the acceleration
of the mass-center of the two bodies while A is sliding on the table? What is
the acceleration of the mass-center after A leaves the table edge? Discuss
the motion of the system after A leaves the table.
i
w
[]
CO
Fig. 374.
Fig. 375.
404. The three bodies A, B, and C (Fig. 375) weigh 6 lb., 4 lb., and 10 lb.,
respectively. They are held in the position shown and then released simul-
taneously. The mass of the pulleys and of the strings is negligible. Locate
the mass-center after the bodies have been moving 2 sec. Find the tensions
in the strings. Ans. y = lS.7 ft.; Ta = 7.S7 lb.
§ 3. Special Topics in Kinetics
153. Introduction. — In this section are discussed several
special problems or topics in kinetics which find direct application
in engineering. The principles and equations of motion employed
in their treatment are developed in the preceding sections of this
chapter and all of the topics here considered could have been
treated as special problems in the preceding sections. However,
the topics are grouped and discussed in this section in order to
give an opportunity to review the principles already developed
and to emphasize their engineering applications.
154. Hoop Tension in Flywheels. — The stress developed in
the rim of a flywheel or pulley, when revolving with a high
HOOP TENSION IN FLYWHEELS
357
angular velocity, is sometimes of considerable importance.
Likewise the stress developed in a belt, due to the same cause, as
it passes around a flywheel or pulley, must sometimes be con-
sidered. If the tension in the
spokes of a flywheel is neg-
lected, the tensile stress in
the rim (often called hoop
tension) corresponding to an
angular velocity, o), may be
found as follows: In Fig. 376
is represented one-half of the
rim of a flywheel. As the
wheel rotates, each half of the
rim tends to separate from
the other half and is prevented
from doing so by the stresses
P, P which are developed in
the rim. The reversed effective force for the half of the rim is
Mro3^ and it acts through the mass-center of the half-rim. And,
since the reversed effective force (inertia force) is in equilibrium
with the external forces (P, P) which act on the half-rim, the
following equation of equilibrium may be written,
2P = Mrco2 = -rco2,
Fig. 376.
in which W is the weight of the half -rim. Now if the thickness of
the rim is small in comparison with the mean radius r, the mass-
center of the rim may be considered to coincide with the centroid
2r
of the semicircular arc, and hence r = — (Art. 82). Whence,
2 g^ir
2r 2 Wrco2
gir
The stress, s, per unit of area of the rim cross-section is
P
in which a is the area of the cross-section. Therefore,
g TT a
>2 kr^co^
358 FORCE, MASg, AND ACCELERATION
in which k is the weight of the material per unit volume. Or, since
the velocity, v, of the mid-points of the rim is equal to cor, the
expression for s may be written in the form,
s= — .
9
The units in which s is expressed are lb. per sq. ft. if A: is expressed
in lb. per cu. ft., r in ft., g in ft. per sec.^, and co in rad. per sec.
It will be noted, therefore, that the intensity of stress, s, developed
in the rim of a rotating wheel, if the rim is thin and the effect of
the spokes is neglected, varies directly as the square of the linear
speed of the rim.
PROBLEMS
405. A common rule limits the peripheral speed of cast-iron flywheels or
pulleys to 6000 ft. per min. (Sometimes stated one mile per minute.) Cal-
culate the tensile unit-stress in the rim corresponding to this epeed, assuming
that the effect of the spokes may be neglected. Ans. 970 lb. per sq. in.
406. Calculate the greatest number of revolutions per minute (r.p.m.)
at which a thin cast-iron hoop can rotate without |Dursting. Assume that the
maximum tensile strength of the cast iron is 20,000 lb. per square inch, that
the material weighs 450 lb. per cubic foot,, and that the radius is 2 ft.
^ 155. Superelevation of Railroad Track. — The wheels of a
locomotive, electric car, etc., when traveling round a curve of
radius, r, on a level track exert a horizontal thrust, H (flange pres-
sure), on the outer rail as shown in Fig. 377(a). The reversed
1^ . .
effective force, Mrcc^ or M- , if acting on the car with the four
r
external forces R\, R2, H, and W, would hold the car in equihbrium.
Hence Mrco^ and H form a couple the moment of which, for a given
car and radius of curvature, depends both upon the speed of the
car and the height of the center of gravity, G, above the rails. In
order to reduce the magnitude of the horizontal thrust and of the
overturning couple, the outer rail is elevated above the inner rail a
distance, e, which is called the superelevation.
It is desired to determine the superelevation required to reduce
the flange pressure to zero for a given speed of the car and curva-
ture of the track. Thus, in Fig. 377(6), the pressures of the rails
are i^i and R2, the resultant of which is R; W is the weight of the
SUPERELEVATION OF RAILROAD TRACK
359
car; and r is the radius of the curve around which the car is trav-
eUng. Since the center of mass of the car travels in a horizontal
plane, the reversed effective force MroP' is horizontal and its action
line passes through the center of gravity, G, of the car, as shown.
Since the reversed effective force is in equilibrium with the external
forces, the three forces W, Mroi^, and R form a concurrent system
in equilibrium. Therefore, we may write,
2i^x = 0, or Rsmd = ^-,
i:Fy = 0, or Rco^e = W.
Mr©-
t
W
R2
(a)
Fig. 377.
And, by dividing the first of these equations by the second, the
resulting equation is,
tan d = —.
gr
Now for small angles the sine and the tangent of the angle are
€
approximately the same. But from Fig. 377(6), sin 6 = -; in which
a
d is the distance between the action lines of the rail pressures
(usually taken as 4.9 ft.). Therefore,
6 iP
tan 6=-^ = —,
d gr'
Hence, if v is expressed in ft. per sec, gr in ft. per sec .2, and d and
r in ft., the superelevation (in ft.) is found from the equation
e =
gr
It will be noted that the value of e which reduces the flange
360
FORCE, MASS, AND ACCELERATION
pressure to zero also reduces the overturning couple to zero regard-
less of the height of the center of gravity above the rails. How-
ever, if a car travels round a curve at a speed greater than that
for which the flange pressure is zero, there will be, of course, an
overturning couple and its magnitude will depend directly upon
the height of the center of gravity of the car and its cargo above
the rails. In order to indicate common values of the superelevation,
the values used on one particular steam railroad are given in
the following table :
SUPERELEVATION OF OUTER RAIL IN INCHES
Degree * of
Curve.
Speed in Miles per Hour.
30
45
60
75
1
t
n
2
3i
2
7
8
21
4
6^-
3
u
3i
6
91
4
If
H
8
—
5
2
5i
—
—
* A one-degree curve is a curve (circle) in which a 100-ft. chord is sub-
tended by a central angle of one degree. In a two-degree curve a chord of
100 ft. is subtended by a central angle of two degrees, and so on.
PROBLEMS
407. The radius of a railroad curve is 1800 ft. What must be the super-
elevation of the outer rail in order to make the flange pressure zero when the
speed of a car around the curve is 50 miles per hour? Ans. e = 5.46 in.
408. Do the superelevations given in the above table reduce the flange
pressure to zero for the speeds specified? Test several values.
Pendulums
156. Simple Circular Pendulum. — A simple circular pendulum
consists of a single particle suspended by a weightless thread
and allowed to swing or vibrate in the arc of a circle, in a
vertical plane, under the influence of gravity and the pull of
SIMPLE CIRCULAR PENDULUM
361
the string. Although these ideal conditions cannot be realized
fully in any physical apparatus, the motion of a small body sus-
pended by a light thread will be approximately that of the ideal
simple pendulum.
It is desired to find the time or period of vibration of a simple
circular pendulum. Let C (Fig. 378) represent a small body which
is suspended by a thread from the point 0 and is allowed to swing
as a simple pendulum in the arc BB' of radius OC or I. By resolv-
ing forces in the direction of the tangent to the path we have,
or,
W
— TF sin ^ = — at,
a
whence,
at=—g^m.d.
And from Art. 127,
vdv=atds.
Therefore,
vdv= —g sin 6 ds.
But,
Therefore,
ds = Idd.
vdv= —Ig sin Odd.
Fig. 378.
The integration of this equation, after expressing v in terms of t
and d, leads to a complicated relation between t and 6. However, if
the angle of vibration is small, a close approximation to the correct
solution may be obtained by assuming that ^ = sin 6. The last
equation then becomes
vdv= —IgSdd.
And by integrating,
To determine, C\, the constant of integration, let v = 0 when
^=^1. Using these values in the above equation, we have,
C, = lgdi\
Hence,
or,
^=Wi^-e^),
t;=\/lgVdi^-0^,
362 FORCE, MASS, AND ACCELERATION
ds
The time, t, may now be introduced in the equation since v = -r.
And, if the particle is moving towards A, s decreases as t increases
ds
and hence -r. is negative. Thus,
-frf=-^^^^^
or,
dd
=P=->
Ve?-
And, by integrating.
If time is measured from the instant the body is at B, then 6 = di
when t = 0. By substituting these values of 6 and t in the last
equation, the value of C2 is found to be sin "^ 1 or -. Thus, the
last equation becomes.
'=VJ(1— ■!)■
By making 6 equal to zero in the last equation, the corresponding
value for t, that is, the time required for the body to move from B
to A is -x^ -. Since the body gains velocity during the displace-
ment BA at the same rate that it loses velocity in the displace-
ment AB', the average velocities for the two displacements are
equal. Therefore, the time required for a single oscillation (B to
B') is IT ^/-. The time or period, P, of a complete oscillation
(from B to B' and back to B) then is,
4
This equation shows that the period P is independent of ^i", that
is, of the amplitude of the oscillation. However, it must be
remembered that the equation applies to oscillations of relatively
small amplitudes.
If the ampHtude of oscillation is not small enough to permit of
COMPOUND PENDULUM
36^
the assumption that sin 0 = 0, then the time of a complete oscilla-
tion is expressed by the series,
P
in which
-'4M^'-^m--m''^ ]
A; = sin
Oi
PROBLEMS
409. A simple pendulum 4 ft. long swings through an angle of 60° (that is,
^1=30°). Find the period of oscillation, (1) by the approximate method and
(2) by the exact method. Ans, (1) P = 2.21 sec; (2) P = 2.25 sec.
410. Find the length of the simple pendulum which beats half-seconds,
that is, one for which the period of a complete oscillation is one second.
411. What is the length of a clock pendulum which will make one beat per
second. If the clock loses 5 sec. per hour, how much should the pendulum be
shortened?
157. Compound Pendultun. — A physical body which oscillates
or swings about a horizontal axis under the influence of gravity
and the reaction of the supporting
axis is called a compound or physi-
cal pendulum.
It is desired to find the time of
oscillation of a compound pendulum.
Let Fig. 379 represent a section of
such a pendulum which rotates about
an axis through 0. From Art. 146
we have as one of the equations of
motion for a rotating body,
Or,
Whence,
2^0 = 70 0!.
— Wf sin d = Ioa
9
(1)
364 FORCE, MASS, AND ACCELERATION
Now, it was shown in the preceding article that the tangential
acceleration of a simple pendulum of length I is,
at= —g sin 6.
But,
at = la (Art. 121).
Whence,
a = y=-^sin^ (2)
By comparing equations (1) and (2) it will be noted that the
angular motion of a simple pendulum may be made exactly the
same as that of a compound pendulum (if the two start from the
same position) by fixing the length I of the simple pendulum such
that,
r
A simple pendulum of this length is called an equivalent simple
pendulum. The period, then, of a complete oscillation, of small
amplitude, of a compound pendulum is.
\ gr
158. Center of Oscillation. — The point, Oi, in the compound
pendulum (Fig. 379) at the distance -^ from the center of rotation
is called the center of oscillation. That is, the center of oscillation
is that point at which the whole mass of the compound pendulum
may be concentrated without changing the period of vibration.
It will be noted that the center of oscillation is also the center of
percussion (see Art. 148).
Further, the center of oscillation may be made the center of
rotation without changing the period of oscillation. That is, in a
compound pendulum the centers of oscillations and suspension
are interchangeable. This fact may be shown as follows: The
distance GOi from the center of gravity of the compound pendu-
lum to the center of oscillation (Fig. 379) is,
r
But, from Art. 102,
A;o2 = A;2+r2.
TORSION PENDULUM
365
Whence,
r
Now if Oi is made the center of rotation, then the new center of
oscillation, O2, will be a distance GO2 = — from the center of gravity.
But r is now equal to GOi ; hence,
P P
002 = 7^ =4r=r.
GOi
Therefore O2 coincides with Oi and hence the centers of sus-
pension and oscillation are interchangeable.
PROBLEMS
412. A uniform slender rod 4 ft. long oscillates as a compound pendulum
about a horizontal axis through one end of the rod, the rod being perpendicular
to the axis, (a) Find the period of oscillation. (6) About
what other point could the rod rotate and have the same /'"NO
period of oscillation. '^
Ans. (a) P = 1.80 sec; (6) 1.33 ft. from end of rod.
413. Find the length of a uniform slender bar having a
period of oscillation of 1 sec. when allowed to swing as
a compound pendulum about an axis through one end
of the bar. Ans. Z = 1.22 ft.
414. A steel pendulum consists of a circular disc 10 in.
in diameter and 1 in. thick, and a rectangular bar 30 in.
long, 3 in. wide, and 1 in. thick, as shown in Fig. 380. If
the pendulum oscillates about a horizontal axis through 0,
what is the period of oscillation?
159. Torsion Pendulum. — A torsion pendulum Fig. 380.
consists of a body (Fig. 381) which is suspended
by a wire or slender rod and allowed to oscillate about the axis
of the wire. The body is rigidly attached to the wire so that
its center of gravity lies on the axis of the wire and the wire is
rigidly attached at the support. Thus, if the disc of Fig. 381
is given an initial angular displacement, ^1, the wire is twisted
and when the disc is released the wire exerts a turning moment on
the disc, which starts it to oscillate.
The torque or turning moment, T, exerted by the wire on the
disc is directly proportional to the angular displacement, provided
366
FORCE, MASS, AND ACCELERATION
that the elastic limit of the material is not exceeded. That is,
T= —kd in which /c is a constant; the minus sign indicates that
when d is positive T is negative.
Since the disc has a motion of rotation,
the equation of motion is,
l^To = Ioa.
Hence,
But,
Therefore,
-ke = Ioa.
df
J-o
The time of an oscillation of the pendu-
lum may now be found by the integration
of this equation. Multiplying each side
of the equation by the factor -r., we have,
dd^^_Jcde
dt df~ h dt'
Whence, by integrating,
Now, when
Therefore,
(f=-i-«-
de
0.
Hence,
-'0
By a method similar to that used in Art. 156, the detailed steps of
which are left to the student, the time of a complete oscillation is
found to be
in which Jo is the moment of inertia of the body (disc) with respect
to the axis of the wire and d is the angular displacement corre-
DETERMINATION OF MOMENT OF INERTIA^ 367
Spending to which the wire exerts the torque T on the disc. It
will be observed that the period of an oscillation is independent
of the initial displacement, that is, of the amplitude of the vibra-
tion, assuming that the elastic limit of the wire is not exceeded.
PROBLEMS
415. The disc of a torsional pendulum is turned through an angle of 10°
by a torque of 1 ft.-lb. When released it is observed to make five complete
oscillations per second. What is the moment of inertia of the disc? (In
using the formula of Art. 159, the angle d must be expressed in radians).
Ans. / = . 00582 slug-ft.2
416. The disc of a torsional pendulum is 1 ft. in diameter and 2 in. thick.
A torque of 1 ft.-lb. is required to turn the disc through 8°. Find the period
of oscillation, assuming that the disc is made of cast iron, the weight of which
is 450 lb. per cubic foot.
160. Experimental Determination of Moment of Inertia. —
As noted in Art. 106, the calculation of the moment of inertia of
many bodies is a difficult process and frequently is impossible.
The moment of inertia, however, may be found experimentally
by allowing the body to oscillate as a compound pendulum and
observing the period of oscillation or by allowing it to oscillate
either as a torsion pendulum itself or with a given torsion pendulum
to which it is attached.
/. By Means of a Compound Pendulum. — The period of a
complete oscillation of a compound pendulum, from Art. 157, is
P
" gr
Or,
\ gr
But,
lQ = Mko^ = -ko^.
g
Hence,
, MP2 _ WP^r
lo=-^gr=^^,
in which 7o is the moment of inertia of the body with respect to the
axis of suspension Now W may be found by weighing the body,
r by balancing the body over a knife edge, and P by observing the
period j)fjDscillation .
368
FORCE, MASS, AND ACCELERATION
After 7o is found, the moment of inertia with respect to a
parallel axis through the mass-center (7) may be calculated from
the equation,
Zo = 7+Mr2.
77. By Means of the Torsional Pendulum. — In order to deter-
mine the moment of inertia of a body by means of a torsional
pendulum, it will be convenient to replace the disc of the pendulum
as described in Art. 159 by two discs rigidly connected as shown in
Fig. 382. If 7o denotes the moment of inertia of the two discs
with respect to the axis of the wire, the time of a complete oscilla-
tion of the pendulum (denoted here for convenience by Po) is, by
Art. 159, _
Po = 2Tyj^-f = 2^^^VTo = CVTo, (1)
o
in which C is a constant. The value of ^ is a constant of the wire,
and depends on the material of the wire and
its diameter. If this value is known (and it is
readily found), 7o may be obtained from the
above equation by observing the period of
oscillation of the pendulum. The method of
determining the moment of inertia of a body
A is as follows: First allow the pendulum to
oscillate and observe the period of oscillation,
Po- Next, place the body A on the lower disc
(Fig. 382) so that the center of gravity of A
is vertically below the wire. If the moment
of inertia of A with respect to the axis of the
wire is denoted by I a and the period of oscil-
lation of the pendulum (and A) is denoted by
P, we have, by Art. 159,
'////m^///^'
Fig. 382.
P = 2x,J(^5d^» = CV/^+77.
(2)
From (1) and (2) the following equation is obtained:
vTo
P
V7o+7/
NEED FOR BALANCING
36d
That is.
Hence,
/o+/a=h^Io«
?1
^A=io{^,-iy
It may be noted that if the discs are of such form that their moment
of inertia may be calculated it is not necessary to know the con-
stant ^ of the wire..
PROBLEMS
417. The connecting rod of a steam engine weighs 300 lb. and the distance
of the center of gravity from the crank-pin is found (by
balancing) to be 50 in. When suspended from the crank-pin
end and allowed to vibrate as a compound pendulum, it is
found to make thirty complete oscillations in 75 seconds.
Determine the moment of inertia of the rod with respect
to the axis of the crank-pin and also with respect to a paral-
lel axis through the center of gravity.
Ans. / = 198slug-ft.2 7 = 36.5 slug-ft.^
418. The pendulum (Fig. 383) of a Charpy impact machine,
which is used to determme the resistance of materials
to impact, weighs 50.5 lb. and the distance of the center
of gravity from the axis of rotation, as determined by
balancing, is found to be 27.33 in. When allowed to vibrate
about the axis of rotation, the pendulum is observed to make Fig. 383.
35 complete oscillations in 61 sec. Find the moment of inertia
of the pendulum with respect to the axis of rotation.
Balancing
161. Need for Balancing. — A moving part of a machine, as a
rule, has either a reciprocating motion similar to that of the cross-
head of a steam engine or a motion of rotation such as that of the
crank-shaft of an automobile engine or the motor of a dynamo.
In any case, if the moving parts are accelerated, forces must be
supplied to produce the accelerations. If the moving parts are
not balanced, the forces which act on the moving masses are
transmitted to them from the stationary parts of the machine such
as the bearings and the machine frame. And, in supplying these
accelerating forces, serious trouble may arise such as, vibrations
in automobiles, turbines, etc.; defective commutations in elec-
370 FORCE, MASS, AND ACCELERATION
trical machinery; heavy bearing pressures which cause undue
wear; defective work with grinding discs, high-speed driUing
machines, etc.; and defective lubrication. It is of great impor-
tance, therefore, to properly neutraUze or balance these forces
in various types of machines.
The moving parts of a machine may be (1) in static or standing
balance or (2) in dynamic or running balance. Standing balance
exists if the forces which act on the parts, when the parts are not
running, are in equihbrium regardless of the positions in which
the parts are placed. Dynamic balancing consists in distributing
the moving masses or in introducing additional masses so that the
inertia forces exerted by the masses of the moving system are in
equihbrium amongst themselves, thereby making it unnecessary
for the stationary parts of the machine to supply any of the
accelerating forces. The complete balancing of a machine, how-
ever, is not always practicable or possible.
The method of balancing rotating masses, only, is here dis-
cussed. For methods of balancing reciprocating masses and for an
excellent discussion of the whole subject of the balancing of engines
see Dalby's " Balancing of Engines."
162. Balancing of Rotating Masses. — The inertia forces or
kinetic loads due to unbalanced rotating masses may be treated,
in general, (1) as a system of centrifugal forces or (2) as a system
of centrifugal couples or (3) as a combination of the two.
IT
I ^^
(a)
Fig. 384.
A Single Rotating Mass. Centrifugal Force. — If a shaft (Fig.
384a) rotates at an angular velocity co and carries a single mass Mi
the center of gravity of which is at the distance ri from the axis
of rotation, the shaft will be subjected to a kinetic load (centrifugal
force) of magnitude Minco^. This kinetic load causes the shaft
BALANCING OF ROTATING MASSES
371
to exert forces on the bearings which in turn are transmitted to
the machine frame. The reactions at the bearings may be elim-
inated by balancing the rotating mass. This may be done by the
addition of a single mass M2 diametrically opposite to Mi (Fig.
3845), the center of gravity of M2 being at a distance r2 from the
axis of rotation, such that,
— ria)^= —
9 9
But, since — is a common factor, the conditions for running or
9
dynamic balance may be expressed by the equation
Mirioy^ = M2r2oP or
co2
2 2
-If a shaft carries
Win = W2r2'
Now, as Is evident from Fig. 384(6), this equation expresses the
condition for standing balance also. Thus, a shop method of
obtaining approximate running balance with a rotating member, in
which the material is substantially in a plane of rotation such as a
disc, a pulley, or a flywheel, consists in drilHng out material on the
heavy side or adding material on the light side until standing bal-
ance is obtained.
Two Rotating Masses. Centrifugal Couple.-
two rotating masses Mi and ikf 2 in
different planes of rotation but in
the same axial plane (Fig. 385)
and, further, if the centrifugal
forces MiTico^ and M2r20)^ of the
two masses are equal, then the
shaft is subjected to an un-
balanced centrifugal couple which ^^^^
is resisted by an equal couple
exerted by the bearings. Or, if
the two rotating masses are to be
balanced, two additional masses,
Ms and M4, must be introduced
in the same axial plane (Fig. 385)
such that the centrifugal couple
Msrso)% or M^VAoy^h which they set up equals the centrifugal couple
of Ml and M2. Hence,
y////.
///7^
777^,
Fig. 385.
Or,
M2r2(^0' = M^r^(ii%,
372
FORCE, MASS, AND ACCELERATION
And, as before, omitting the common factor — we may write,
y
Wiria = W3r3h, etc.
It will be noted that the shaft when carrying only the two masses
Ml and M2 is in standing balance but not in running balance.
163. Several Masses in a Single Plane of Rotation. — If sev-
eral masses Mi, M2, Ms, etc., lie in the same transverse plane (Fig.
386a), the shaft is subjected to the centrifugal forces,
MinoP, M2r2co^, Msrsoi^, etc.,
which form a concurrent system of forces.
The condition that such a force system shall balance is that the
force polygon shall close. That is, the vectors representing the
forces Mirico^, ^2^2^^, Msrsco^, etc., if laid off in succession, each in
Fig. 386.
its proper direction, shall form a closed polygon. Or, since MroP
W oP
may be written — rco^ and since — is a factor common to the expres-
«7 y
sion for each force, the products Win, W2r2, etc., may be used
instead of the actual forces. Thus, let the four masses as shown
in Fig. 386(a) be a system of masses which rotate in a transverse
plane. Suppose that the products WiVi, W2r2, Wsvs, and 174^4
when laid off as vectors (Fig. 3866) do not form a closed polygon.
It is evident then that the four masses are not in running balance.
In order to balance the system of masses, a mass. Mo, must be added
at a distance, ro, such that the product Woro is represented both in
MASSES IN DIFFERENT TRANSVERSE PLANES
373
magnitude and direction by the closing side, EA, of the vector
polygon. By assuming a convenient value for ro, a value of Mo
may be found. The gap, EA, may be closed, however, by two
or more vectors from which two or more masses may be found that
will balance the system.
164. Masses in Different Transverse Planes. — In Fig. 387,
let the masses Mi and M2 be connected with the shaft at A and B,
respectively, and through some point, 0, of the shaft let a
Fig. 387.
transverse plane, called a reference plane, be chosen. The mass Mi
exerts a kinetic load or centrifugal force, Fi, on the shaft such that
/^i = Mirico2.
Now at 0 introduce two equal and opposite forces each equal and
parallel to Fi. The force Fi at A and the equal opposite force at 0
form a couple, Ci, the moment of which is
Ci = Mirio3^ai
Thus the single force Fi at A is replaced by an equal parallel force
at 0 and the couple Ci . In hke manner the single force F2 = M2r2oy^
at B may be replaced by the equal parallel force F2 in the reference
plane and a couple C 2 = M2r 2(^^(12-
Thus, the kinetic loads on a shaft exerted by a system of
rotating masses may be reduced to a system of concurrent forces
374
FORCE, MASS, AND ACCELERATION
in an arbitrarily chosen reference plane and a system of couples
which he in different axial planes. The resultant of the system
of concurrent forces, if not balanced, is a single force in the refer-
ence plane, and the resultant of the system of couples, if not bal-
anced, is a single couple in some axial plane. Hence, in general,
the system of kinetic loads, if not balanced, may be reduced to a
single force and a couple. The magnitude of the couple will, of
course, depend on the position chosen for the reference plane.
The conditions then which must be fulfilled to have a system of
rotating masses in equilibrium are,
(1) The resultant of the system of concurrent forces must
be zero. That is, the force polygon for the forces in the reference
plane must close.
(2) The resultant of the system of couples must be zero.
That is, the couple polygon must close.
These conditions may be satisfied by the addition of two rotating
masses in different transverse planes. Thus, let the shaft (Fig.
388) carry an unbalanced system of rotating masses, ilf i, M2, and
Fig. 388.
Ms, and let the two balancing masses be denoted by Mo and ilf' 0.
Let the transverse planes in which these balancing masses are to lie
be chosen arbitrarily but let the plane of one of the masses be chosen
MASSES IN DIFFERENT TRANSVERSE PLANES 175_^
as the reference plane. The plane of mass iW'o will here be selected
as the reference plane. Let oo, ai, a2, and as denote the respective
distances of the masses Mo, Mi, M2 and 71^3 from the reference
plane. The couples Wiviai, W2r2a2, etc., may now be calcu-
lated (the common factor — is omitted in each term for simplicity).
The only unknown couple is TTo^o^o, since the couple W'or'oa'o
is zero due to the fact that the reference plane was chosen as the
plane of M'o, which makes a'o equal to zero. By laying off the
vectors that represent the known couples as the sides of a polygon,
the closing side determines both the magnitude of the unknown
couple WoVoOo and the axial plane in which it lies. By assigning
any convenient value to the moment arm oo, the product Woro
may be found, and by assuming a convenient value for ro, TFo may
be determined and placed in the plane indicated by the closing
vector of the couple polygon. Thus, by the addition of the
couple Wovocio condition (2) is satisfied.
Condition (1) may now be satisfied as follows: The kinetic
load due to the added mass Mo is replaced by a force in the refer-
ence plane and the couple Morooo as was done for the kinetic loads
due to the original masses. Now, if the products Wiri, W2r2,
etc. (including T^o^o) are laid off as the sides of a polygon, the
closing side gives the magnitude and the direction of a product
W'or'o for a body Wo which must be added in the reference plane
to balance the system of concurrent forces in the reference plane
and thereby satisfy condition (1). By choosing a convenient
value for r'o, the mass M'o may be determined and the direction
of the closing side of the polygon gives the direction of M'o from
the axis of the shaft.
If the couple polygon is formed by drawing the couple vectors
perpendicular to the planes of the couples, as explained in Art. 17,
and then is turned through 90°, it will be the same as the polygon
formed by drawing the couple vectors according to the following
rule: Draw the couple vectors parallel to the respective crank
directions; outwards for masses on one side of the reference plane;
inwards towards the axis for masses on the opposite side of the
reference plane.
The vectors of the force polygon are drawn, of course, from the
axis outwards parallel to the radii to the masses.
376
FORCE, MASS, AND ACCELERATION
ILLUSTRATIVE PROBLEM
419. Three weights Wi, W2, and TF3 (Fig. 389) which revolve in the planes
1, 2, and 3 are to be balanced by the addition of two weights. Plane 1 is
chosen as the plane of one of the weights (Wo) and the plane of the other
weight (W'o) will arbitrarily be taken 1.4 ft. to the right of plane 3 and will be
selected as the reference plane. It is required to determine values of T^o and
W'o and the lengths and directions of the corresponding radii for kinetic
balance. The accompanying table gives the values of the weights, the lengths
and directions of the radii, and the distances from the reference plane.
Plane.
Weight
W (lb.)
Radius
r(in.)
Angle e.
Distance
from R.P.
a (ft.)
Wr (Ib.-in.)
Wra
(Ib.-in.-ft.)
1
2
3
0
0' (R.P.)
2
3
3
(1.36)
(2.28)
9
8
6
(8)
(5)
270°
30°
150°
4.4
2.6
1.4
4.4
0
18
24
18
(10.9)
(11.4)
79.2
62.4
25.2
(48.0)
Solution. — From the given data, the values of the products Wr and Wra
for the known weights are calculated and entered in the last two columns of
the above table. The couple polygon is then drawn as shown in Fig. 389(c).
AB is laid off in the direction of Wi outwards from the shaft and its length
represents to scale the value of the product Wiriai which is 79.2. Next BC is
laid off in the direction of W2 such that its length represents to the same scale
Wir-itti (62.4) . Similarly CD is laid off to represent Wsr^as (25.2) . The closing
side DA of the polygon represents the product Woroao due to the balancing
weight Wo in plane 1 (or O). This product is found by measuring to be 48.0.
Hence,
Troro=^-,^ = 10.9 1b.-in.
4.4
This product is now entered in the column of the above table with the other
TTr-products. The addition of the weight Wo in the plane 1 at the distance ro
reduces the resultant couple to zero. There are left, however, the forces
in the reference plane, including the force corresponding to the product WoTq
just found, and these forces will, in general, not be balanced. Now if the
vectors corresponding to these Tfr-products are laid off in order as in Fig.
389 (d), the closing vector EA represents the product W'or'o which by measure-
ment is found to be 11.4. If the values of If oro and W'or'o are divided by
assumed values of ro and ro' (8 in. and 5 in. respectively) we obtain the values,
Tf 0 = 1.36 and T^o' = 2.28. Hence if weights of 1.36 lb. and 2.28 lb. are placed
MASSES IN DIFFERENT TRANSVERSE PLANES
377
in planes 0 and O' at radial distances 8 in. and 5 in., respectively, as indicated
in Fig. 389, the system will be in kinetic balance.
Fig. 389.
PROBLEMS
420. Five bodies are attached to a disc which revolves with constant
angular velocity. In the following table are given the weights of the bodies,
the angles which the radii from the mass-centers to the axis of rotation make
with the X-axis, and the lengths of the radii to the mass-centers. The angles, 6,
are measured counterclockwise from the x-axis.
w
d
r
5 1b.
45°
18 in.
6 1b.
120°
15 in.
10 1b.
150°
12 in.
4 1b.
240°
12 in.
6 1b.
315°
18 in.
378
FORCE, MASS, AND ACCELERATION
Determine sufRcient data for kinetic balance of the system of bodies:
(a) By the addition of a single weight placed 2 ft. from the axis of
rotation.
(6) By the addition of two weights, one of 2 lb. placed on the y-axis,
and the other of 2.25 lb. placed 2 ft. from the axis of rotation.
421. Four bodies are attached to a revolving shaft in different transverse
planes. The weights and positions of the bodies are indicated in the following
table, the reference plane being the transverse plane in which the mass-center
of the 6-lb. body (W^) Hes.
w
d
.
a
5 1b.
30°
2 ft.
-1 ft.
4 1b.
45°
lift.
2 ft.
6 1b.
150°
1 ft.
0 ft.
4 1b.
240°
2 ft.
hit.
The given masses are to be balanced by two masses, one mass to be placed
in plane 4 at a radial distance of 1| ft., and the other to be placed in the
reference plane at a radial distance of 2 ft. Find the weights and values of
d for the two masses. Ans. PFo = 2.93 lb.; W'o = 4:.85 lb,
422. The crank shaft of a gas engine carries two flywheels A and B, the
planes of revolution of which are 3.5 ft. apart. The plane of revolution of the
crank is between the flywheels and 1 ft. 7 in. from the plane of A. The crank
arms and pin are equivalent to a weight of 108 lb. at a radial distance of 10 in.
from the crank shaft and in the plane of revolution of the crank. What
weights placed at a radial distance of 2 ft., one in each flywheel, will balance
the crank? Solve algebraically. Ans. Pf^ =24.6 lb.; 1^5 = 20.4 lb.
The Gyroscope
165. The Problem Defined. — Gyroscopic motion occurs when-
ever a body rotates about an axis in the body as the axis (and
body) is turned about a second axis. Thus, the wheels of a loco-
motive when rounding a curve, or the screw propeller of a ship
when the ship is pitching in a rough sea, are given gyroscopic
motion. The forces that act on the body in giving it gyroscopic
motion may be of considerable importance since under certain
conditions they are very undesirable as, for example, the forces
exerted on the propeller of an aeroplane when making a sharp
turn. On the other hand the gjrroscope is sometimes used to
introduce desirable forces as, for example, in reducing the rolling
of ships.
FORCES IN THE GYROSCOPE
379
The gyroscope here considered is a body symmetrical with
respect to each of three rectangular axes. The body rotates or
spins with constant angular velocity about one of the axes and at
the same time turns about one of the other axes with constant
angular velocity. Thus, in Fig. 390, let A BCD represent a circular
disc which is symmetrical with respect to the three axes x, y, and z.
Let the disc rotate or spin with a high constant angular velocity co
Fig. 390.
about the 2;-axis and at the same time let the disc turn about the
2/-axis with the constant angular velocity 12. The problem which
arises is that of determining the forces which must act on the disc
or its axles (axle reactions) in order to maintain this gyroscopic
motion.
166. Analysis of Forces in the Gyroscope. — Due to the two
rotations imposed on the disc, a particle, m, at the distance p
(assumed at the circumference of the disc for convenience)
has at any instant, two velocities: (1) a constant velocity cop in the
plane of the disc due to the rotation about the 2-axis with angular
380 FORCE, MASS, AND ACCELERATION
velocity co, and (2) the velocity 12p cos 6 perpendicular to the disc
(parallel to the z-axis) due to the rotation about the 2/-axis with
angular velocity 12.
The changes that occur in each of these two velocities of any
particle due to each of the rotations which are given to the disc
will first be investigated, since the effective force for any particle
must have components corresponding to the change in the mag-
nitude and to the change in the direction of each of these two
velocities (Theorems I and II, Art. 118, and Newton's second law,
Art. 140). And, the effective forces for the particles must be
found in order to determine the external forces required to main-
tain the gyroscopic motion. (Read Art. 144 for the general method
of procedure, keeping in mind that gyroscopic motion is not uni-
planar motion.) It will be found, however, that external forces
come into action as the result of only two of the changes which
occur in the two velocities; namely, the change in Up cos 6 due to
the rotation about the 2;-axis and the change in cop due to the
rotation about the ^-axis.
Changes in the Velocities Due to Rotation about Z-axis
(1) Change in cop. — Since co is constant, cop changes in direction
only. The resulting acceleration (co^p) and the corresponding
effective force for the particle, is directed towards the center of
rotation (towards the z-axis). And, since the body is symmetrical,
these effective forces for all the particles form a balanced system
and hence, according to D'Alembert's principle, no external forces
act on the body by reason of this change in the velocity of the
particles.
(2) Change in 12p cos 6. — In one revolution of the disc about the
2:-axis, this component of the velocity changes, as follows: It
gradually increases in magnitude downwards from zero at A to a
maximum at 5; it then decreases gradually to zero at C; then
increases in the opposite direction to a maximum at D] and finally
decreases to zero again at A. The acceleration, ai, at any instant
then is (Theorem I, Art. 118),
a\ = -T. = — 12p sin ^-^ = — 12p sin ^co = — col22/,
in which y is the distance of the particle from the x-axis, and the
FORCES IN THE GYROSCOPE 381
minus sign indicates that the sense of the acceleration is opposite
to that of the velocity for the position of the particle as shown in
Fig. 390. Hence, for the position of the particle shown, the direc-
tion of cofl^/ is upwards. The effective force, corresponding to this
acceleration, for any particle in the quadrants OBC and OCD, is an
upward force perpendicular to the plane of the disc. And, in the
quadrants ODA and OAB, it is a downward force. The resultant of
the effective forces for the four quadrants, then, may be represented
by the forces P (Fig. 390). These forces form two couples which
have moments with respect to only one of the rectangular axes;
namely, the a:-axis. Further, according to D'Alembert's principle,
the effective forces require that external forces act on the disc
such that the resultant of the external forces is equivalent to that
of the effective forces. Hence, an external couple must act on the
disc (or its axles) as indicated by the forces Q (Fig. 390).
Changes in Velocities Due to Rotation about Y-axis
(3). Change in o)p. — Let the velocity cop be resolved into two
components; one parallel and one perpendicular, respectively, to
the 2/-axis (Fig. 390). The component parallel to the 2/-axis under-
goes no changes due to the rotation about the ^/-axis; but the
component cop sin 6, which is perpendicular to the 2/-axis, changes
in direction due to the rotation with angular velocity 12 about the
y-Sixis. And, according to Theorem II of Art. 118, the accelera-
tion a2 corresponding to this change in velocity is the product
of the magnitude of the velocity and its angular velocity of turn-
ing, that is,
a2 = o)p sin ^ • 12 = col2?/,
and it is directed upwards perpendicular to the plane of the disc.
This acceleration, therefore, is equal to ai (see above) and has the
same direction and sense as that of ai for all positions of the par-
ticle, as will be observed from a study of Fig. 390. Therefore,
external forces which act on the disc as a consequence of the
changes in cop sin 6 due to the rotation of the disc about the y-axis
must constitute a couple exactly the same as that found under (2).
(4) Change in 12p cos 6. — This velocity changes in direction,
only, due to the rotation about the y-axis with constant angular
velocity 12. The resulting acceleration, fi^Xp cos 6 or 12^0;, is
382 FORCE, MASS, AND ACCELERATION
directed towards the center, on the y-axm, about which the par-
ticle is rotating at the instant. The corresponding effective forces
for the particles of the whole disc, therefore, form a parallel system
in the plane of the disc. But, since the disc is symmetrical with
respect to the ^-axis, this effective force system is balanced and
hence no external forces act on the disc as a consequence of the
changes caused in this velocity by the rotation about the ^/-axis.
It will be noted, therefore, that if the disc is rotated about
the 2;-axis and at the same time is turned about the y-axis, it will
rotate about the a;-axis unless an external couple acts on the disc to
prevent the rotation. This external couple is called the gyroscopic
couple.
A simple experiment for demonstrating the existence of the
gyroscopic couple may be made by holding a bicycle wheel (dis-
mounted from the frame) with one hand on either end of the pro-
jecting (horizontal) axle. If the wheel is spinning in the vertical
plane about the axle which is held in the hands, any attempt to
turn the axle (and hands) in the horizontal plane wiU cause the
wheel (and hands) to turn about a horizontal axis perpendicular
to the axis of the wheel unless the hands exert a couple to prevent
this rotation.
167. The Moment of the G3rroscopic Couple. — In the preceding
article it was shown that the only acceleration of a particle of
the disc which requires the action of external forces on the disc is
a = ai-\-a2 = o)^y-{-coQy
= 2o)^y,
and that it is directed perpendicular to the plane of the disc;
upwards in the two quadrants BCD, and downwards in the two
quadrants DAB.
The force required to produce this acceleration (effective force)
of the particle of mass m is,
F = ma = mX 2co%,
the direction of which agrees with that of a.
The moment of this effective force for the particle, about the
a;-axis is,
FXy==2mo)Qy^,
MOMENT OF GYROSCOPIC COUPLE 383
and the sum of the moments of the effective forces for all the par-
ticles of the disc is,
S2mco%2.
But since co and 12 are constant this may be written,
26ol2Smi/2 = 2/xCol2,
in which Ix is the moment of inertia of the .disc with respect to the
X-axis.
Now the sum of the moments of the effective forces is equal to
the moment of their resultant, but the resultant of the effective
forces is a couple, as shown in the preceding article. Further, this
resultant couple is equal to the external or gyroscopic couple.
Therefore, the moment, C, of the gyroscopic couple is
C = 2/xcol2.
But, since the disc is symmetrical with respect to the x- and i/-axes,
Ix is equal to ly. Hence, by making use of the equation of Art. 89,
2Ix may be replaced by the moment of inertia of the disc with
respect to the axis of spin (2;-axis). Therefore, the moment of the
gyroscopic couple is,
in which I is the moment of inertia of the disc with respect to the
axis of spin.
The following conclusion then may be drawn : If a body is sym-
metrical with respect to each of two rectangular axes {x and y)
and rotates or spins with a constant angular velocity co about a
third axis perpendicular to each of the two axes (the 2;-axis or axis
of spin), a couple having a moment about one of the two axes
(the a:-axis) is required to maintain an angular velocity, 12, about
the other of the two axes (the 2/-axis) ; the moment of the couple
is equal to the product of (1) the moment of inertia, 7, of the body
with respect to the axis of spin (2;-axis), (2) the angular velocity,
w, of spin, and (3) the angular velocity, 0, about the ^/-axis.
The angular velocity 12 which is maintained by the couple
is called the velocity of precession and the corresponding axis
(2/-axis) is called the axis of precession. The axis about which the
384
FORCE, MASS, AND ACCELERATION
couple Icofl tends to rotate the disc (a:-axis) is called the torque
axis. Hence the disc when spinning about the 2:-axis with angular
velocity co is said to precess about the y-axis when acted on by a
couple having a moment of /wfi about the x-axis.
By referring to Fig. 390, it will be seen that the sense of rotation
about the axis of precession (y-Bxis) is in accordance with the fol-
owing rule:
The sense of precession is such as to turn the axis of spin
toward the torque axis, that is, the axis of spin tends to become
coincident with the torque axis.
In the interpretation of this rule the torque axis must be
regarded as that part or end of the a:-axis which, considered as a
vector drawn outward from the origin, represents the moment or
torque of the couple lojfi, and the axis of spin must be regarded
as that part or end of the z-axis which, considered as a vector drawn
outward from the origin, represents the angular velocity co about
the 2;-axis.
Thus, if a disc of weight W (Fig. 391) is given an angular
velocity w about the 2;-axis and
then one end of the axis is placed
on the vertical post at A, the
couple having a moment Wl will
cause the disc (and z-Sixm) to ro-
tate with angular velocity 12
about the axis of the post (^/-axis)
such that
Or,
^__ Wl gl
Fig. 391. IF., ^^^'
in which k is the radius of gyration of the disc with respect to the
axis of spin (2:-axis). Further, if the sense of oi be as represented
in Fig. 391, the sense of precession about the 2/-axis will be clock-
wise as viewed from the positive end of the y-axis. For a further
discussion of the gyroscopic couple see Art. 221.
MOMENT OF GYROSCOPIC COUPLE
385
ILLUSTRATIVE PROBLEM
423. The flywheel of an engine on a ship weighs 6000 lb. and has a radius
of gyration of 3.75 ft. It is mounted on a horizontal axle which is parallel to
the longitudinal axis of the ship, and has a speed of 400 r.p.m. clockwise when
viewed from the rear. Find the gyroscopic couple when the ship is turning
to the left with an angular velocity of 0.1 rad. per sec. What are the axle
reactions if the distance between the centers of bearings is 4 ft.?
Solution. — The moment of inertia of the flyivheel about the axis of spin is,
And,
/=^X (3.75)2 = 2620 slug-ft.^
co = — 7^ — =41.9 rad./sec.
60
Hence the gyroscopic couple is,
/o>12 = 2620X41.9X0.1 = 10,980 Ib.-ft.
In accordance with the rule stated in Art. 167, the vector representing this
couple is perpendicular to the axis of the ship and is directed towards the right.
The forces constituting the gyroscopic couple are the axle reactions and hence
the reaction at the forward bearing is downwards and that at the rear bearing
is upwards. Since the distance between centers of bearings is 4 ft. the magni-
tude of each of these reactions is 10,980 -7-4 = 2745 lb. The effect of the
gj-Toscopic motion, then, is to increase the reaction at the rear bearing and to
decrease it at the forward bearing. The reaction at each bearing due to the
weight of the flywheel is 3000 lb. Hence the resultant reaction at the rear
bearing is 3000+2745 = 5745 lb. and that at the forward bearing is
3000-2745 = 255 lb.
PROBLEMS
424. A disc 4 ft. in diameter (Fig. 392) rolls on a circular track having a
radius of 10 ft. The center of the disc has a velocity of 20 ft. per sec. The
Fig. 392.
386 FORCE, MASS, AND ACCELERATION
disc is attached to the central axis OF by means of a rod which is coUinear
with the 2-axis about which the disc turns. The disc has a flange similar to
that on a car wheel. If the weight of the disc is 450 lb. find the tension, T,
in the rod and the pressure, P, of the track against the flange of the wheel.
425. A circular disc is mounted on a horizontal axle which is free to rotate
about a vertical axis as shown in Fig. 391, the distance from the center of the
disc to the vertical axis being 2 ft. The radius of the disc is 6 in. and its weight
is 10 lb. If the disc rotates about the horizontal axle with a speed of 300
r.p.m., with what velocity will it rotate about the vertical axis?
Ans. 9. = 157 r.p.m.
426. The propellor of an aeroplane rotates clockwise when viewed from the
rear. If the aeroplane turns to the right when moving in a horizontal plane,
what will be the effect of the gyroscopic couple?
427. The flywheel of an automobile engine is mounted on a horizontal axle
parallel to the longitudinal axis of the automobile. The flywheel rotates
counter-clockwise when viewed from the rear. What will be the effect of the
gyroscopic couple on (a) the axle reactions of the fljrvvheel, (6) the pressures
of the wheels on the road?
Governors
168. The Action of Governors. — The governor of a steam
engine, hydraulic turbine, gas engine, or other motor, automatically
regulates the supply of the steam, water, gas, or other fluid, so
as to keep the driving force exerted by the working fluid constantly
adjusted to the resistance to be overcome. The governor partakes
of the motion of the motor so that an increase in the speed of the
motor due to a decrease in the load, causes a corresponding increase
in the speed of the moving parts of the governor which, in turn,
causes, by means of a suitable mechanism, either a decrease in
the pressure of the fluid or a decrease in the quantity of fluid dehv-
ered to the motor.
The forces which cause the adjustment of the controlling valve
and which are brought into play by the change of motion of the
governor parts, depend in the main upon (1) the actual change
in the speed of the moving parts or (2) the rate of change of the
speed of the moving parts. Governors in which the governing
action depends mainly upon the actual change of speed of the gov-
ernor parts are called centrifugal governors and may be either jpen-
dulum or shaft governors. Governors in which the governing
action depends mainly upon the rate of change of speed of the
moving parts are called inertia governors and are shaft governors.
THE CONICAL PENDULUM
387
A brief analysis of the forces brought into play by a change in the
motion of each of these types of governors will here be made.
169. The Conical Pendulum.—
In Fig. 393(a) a ball of mass M and
weight W is held at the end of an
arm AB and the ball and arm are
caused to rotate about an axle AO
with velocity co. The acceleration
of the center of the ball is directed
towards, 0, the center of its circular
path and its magnitude is rco^.
Hence, the force, R, required to
produce this acceleration of the
ball is,
W
9
Now, the accelerating (effective)
force R must be the resultant of
the external forces which act on the ball. These external forces are
the weight, W, and the tension, T, in the arm, as shown in Fig.
393(6). If a force which is equal but opposed to R (the inertia
force) is assumed to act with W and T as shown in Fig. 393(6), the
three forces will be in equilibrium and hence the sum of their
moments with respect to the point A is equal to zero. Thus,
W
Wr=—n^^Xh.
9
Mrw2
Fig. 393.
Therefore,
'^=S
in which co is expressed in radians per second and g is equal to 32.2
ft. per sec.2 If the number of revolutions per minute (r.p.m.)
is denoted by n, then co =
(in inches) by the equation.
2Trn
"60 •
Therefore, h, may be expressed
35200
n'
(1)
This equation shows that the height h of the cone depends only
on the speed of rotation and not upon the weight of the ball nor
the length of the arm AB. Now, in the pendulmn governor, the
governing action depends upon the manner in which h varies with
388
FORCE. MASS, AND ACCELERATION
CO. The accompanying table is obtained from equation (1). It
will be noted that at low speeds a small change in speed of the ball
n (in r.p.m.)
h (in inches).
20
88.0
40
22.0
60
9.8
80
5.5
100
3.5
150
1.56
200
.88
300
.39
causes a large change in the height
h while at high speed a large change
in speed causes only a small change
in h. If the conical pendulum is used
as a governor as in the simple Watt
governor (Fig. 394) it may be made
sensitive to small changes in speed
A
Fig. 394.
either by making h large or by
causing the governor to rotate
slowly by gearing it down. How-
ever, a decrease in the speed of
the governor results in a decrease
in the energy available to over-
come the resistances of the valve
which is operated by the governor.
In order to make the governor
sensitive to small changes in
speed without seriously reducing
its available energy it may be
loaded.
170. The Loaded Governor.
— The height h of the conical
pendulum may be increased by loading the balls as shown in Fig.
395(a). The disc, D, which rests on the balls (or cylinders) fur-
nishes, by its weight, a vertical external force without influencing
THE PORTER GOVERNOR 380
the inertia force of the ball or cylinder. If L denotes the weight
of the disc (load) then ^ is the load on each ball and the external
forces W, o , and T which act on the ball form, with the reversed
effective (inertia) force Mroy^, a system in equilibrium as shown
in Fig. 395(6). The sum of their moments, therefore, about the
point A is equal to zero. Thus,
{w^^r = Mro?Xh,
and by expressing w in revolutions per minute, h is expressed in
inches by the equation,
,^_ ^2^35200
This equation shows that the addition of the load to the balls of
L
the conical pendulum increases the height h in the ratio of l+o^
to 1. For example, if TF = 5 lb. and L = 30 lb. then,
..L , , 30 .
1+2F=1+2X5 = ^-
Thus, the height h of the simple conical pendulum corresponding to
a speed of 200 r.p.m. is 0.88 in. and when loaded its height becomes
4X0.88 = 3.52 in.
For high rotative speeds, loading must be resorted to in order
to increase the change of height for a given change in speed, that is,
in order to increase the sensitiveness of the governor.
171. The Porter Governor.— In Fig. 396(a) is shown the Porter
loaded governor, in which the load on the conical pendulum is
suspended from additional links which are attached to the balls.
With this arrangement the load is twice as effective as in the loaded
governor discussed in the preceding article, since, for a given rise
of the balls, the load rises twice the distance that the balls rise.
If, therefore, L denotes the load, the expression for h (in inches)
becomes,
TF+L 35200
390
FORCE, MASS, AND ACCELERATION
Thus, in the above example, for the Porter governor, the height
becomes
I^X.88 = ^°X0.88 = 6.16i„.
W o
MFcy
Fig. 396.
The student should derive the above equation for h from the
analysis of the forces in a manner similar to that used in the pre-
ceding two articles. Free-body diagrams
for the revolving ball and load are shown
in Fig. 396(6).
PROBLEM
428. A spnng-loaded governor is shown in Fig.
397. Let W be the weight (lb.) of each ball; r (ft.)
the radius of the path of the balls; I (ft.) the length
of each of the four arms; w the angular velocity in
radians per second. When the radius, r, is zero,
the tension in the spring is T \h. and the force re-
quired to elongate the spring a unit length is Q lb.
Show that, if the weight of the balls be neglected,
., g[T+2Qil-VP^')]
03^ = . — .
If W = S lb., 1 = 1 ft., and the balls revolve at 26
THE CENTRIFUGAL SHAFT GOVERNOR
391
Mrcos
Fig. 398.
rad. per sec. when r = 3 in., find the tension in the spring, assuming the
modulus of the spring to be 58 lb. per in. Ans. T = 16.8 lb.
172. The Centrifugal Shaft Governor. — In the pendulum gov-
ernors considered above, the governing action is dependent upon
the centrifugal force (inertia force) of a rotating mass and the same
is true for some types of shaft
governors. Thus, let a body
of mass M be pivoted to the
arm of a flywheel as shown in
Fig. 398. Let the distance of
the center of the mass, G, from
the shaft center, >S, be denoted
by R and let the wheel rotate
with an angular velocity w.
The centrifugal or inertia force,
then, of the mass M is MnJ^.
As long as co remains constant
this force has a definite value.
To hold it in equilibrium a spring is employed. The moment of the
spring tension about the point 0 must therefore be equal to the
moment of MroP' about the same point (the weight of the body is
small in comparison with MroP' and hence its moment is neglected).
If now the angular velocity of the wheel increases to w' due to a
decrease in the load on the engine, assuming the value of r to
remain constant, the centrifugal force increases to Mroj'^ and the
excess force Mr{o)'^ — o)^) causes the mass to move outwards whicj^
in turn changes the time of cut off of the valve and adjusts the
amount of the working fluid deliv-
ered to the engine.
PROBLEMS
429. What must be the initial tension
in the spring and the modulus of the
spring so that the weight, A, in the gover-
nor shown in Fig. 399 will not leave the
inner stop and pass to the outer stop until
a speed of 210 r.p.m. is reached? Assume
that the weight of J. is 16.1 lb., a = 3 in.,
6 = 24 in., c = 8 in., and d = lQ in. If the
spring should be placed farther from E
than 16 in., would A reach the outer
Fig. 399.
392 FORCE, MASS, AND ACCELERATION
stop with the wheel revolving at the same speed that it had when A left
the inner stop? Ans. T = 242 lb.; modulus =45.4 lb. per in.
430. Given the arrangement as shown in the previous problem, with the fol-
lowing data : a = 3 in., 6 = 20 in., c = 9 in., c? = 12 in., modulus of spring = 50 in. If
the mean speed, no, is 200 r.p.m., what must be the weight of A for a
coefficient of steadiness, , of 0.01, where ni and n2 denote the maximum
no
and minimum speeds respectively? Ans. Ty = 14.85 1b.
173. The Inertia Shaft Governor. — As already noted, governors
in which the regulating action depends upon a change in the cen-
trifugal force, whether of the pendulum or of the shaft type, are
called centrifugal governors. It
is important to note that with
centrifugal governors there must
be an actual change of speed to
give a governing action. In the
inertia governor, however, the
governing action is entirely differ-
ent. Thus, m Fig. 400, let a
mass M be pivoted to the arm
of a flywheel so that the pin, 0,
passes through the center of mass.
The centrifugal force of the mass
is balanced by the pin reaction at
0 and hence the centrifugal force is not involved in the governing
action.
In order to show how the forces which cause the governing
action arise, let it be assumed first that the wheel of Fig. 400 is
standing still and that the mass is turned on the pin, 0, with an
angular acceleration a. To produce this acceleration a moment, jT,
is required having a magnitude,
T = Ioa,
in which 7o is the moment of inertia of the mass with respect to
the pin 0. Now, if a change in speed of the wheel (and shaft)
occurs, due to a change in the load on the motor, each particle in
the mass M, by virtue of its inertia, will tend to maintain its linear
velocity and hence will turn relative to the wheel about the axis 0
and thereby cause the valve mechanism to exert forces on the mass,
COMPARISON OF TWO TYPES OF GOVERNORS 393
the moment of which about 0 will be equal to ha. In this type of
governor, therefore, the governing action depends not upon the
actual change of speed but upon the rate of change of angular
speed, that is, upon the angular acceleration of the wheel or engine
shaft. Shaft governors of this type are called inertia governors.
174. Comparison of the Two Types of Governors. — Centrifugal
and inertia governors differ in two important particulars. Since,
in the centrifugal type, the governing force depends upon the
change in the speed there must be an increase (or decrease) in the
speed of a definite amount before the governing force is sufficient
to move the valve gear against the frictional forces. In the inertia
governor, on the other hand, the governing force is proportional to
the angular acceleration and, in general, the angular acceleration of
the governing mass is greatest just at the beginning of the change
of speed; hence, before the speed of the shaft has changed appre-
ciably, the governor begins to act although the change in the
centrifugal force is small. It follows then that the inertia gov-
ernor acts more quickly and holds the speed within much smaller
limits than does the centrifugal governor.
The centrifugal type of governor, however, has one essential
property not possessed by the inertia type. The centrifugal
force Mro)^ varies somewhat for different positions of the balls in
Figs. 393, 394, 395, or 396, and of the body in Fig. 398, but for any
fixed position it must have a fixed definite value which is deter-
mined by the weight of the balls (and load on the balls) in the pen-
dulum governors and by the spring tension in the shaft governor.
It follows that, with the governor in a definite position, the speed,
CO, of the shaft has a definite value and therefore the speed of the
engine is fixed. Thus, if it is determined that the speed shall be
200 r.p.m. and the governor is properly adjusted, the speed cannot
vary much from that speed as long as the governor is operative.
The speed may sink to 195 r.p.m. under a heavy load or rise to
205 r.p.m. with a light load; but the centrifugal force imposes the
average speed of 200 r.p.m., and the engine cannot be made to
run at a different average speed without some adjustment of the
governor, such as the addition of a weight or an increase in the
spring tension.
175. Analysis of Forces in the Rite*s Inertia Governor. —
In the shaft governor known as the Rite's inertia governor, a
combination of the two governing actions discussed above is
394
FORCE, MASS, AND ACCELERATION
affected by means of a single heavy mass which is pivoted at a
point at some distance from the shaft center but not at the center
of the mass.
To gain a clear conception of the forces which are developed
during the action of a shaft governor of the Rite's type, consider
the motion of a heavy mass of any form (Fig. 401a) pivoted at a
point 0 which is at a distance e from the shaft center S. The
Fig. 401.
center of gravity, (r, of the mass is at the fixed distance s from the
pivot 0 and at the distance r from the shaft center *S; this latter
distance may of course vary. Let any particle, P, of the mass be
at the distances a and p from 0 and S, respectively. The procedure
followed in the analysis of the forces in the Rite's governor is the
same as that used in the analysis of the various kinetics problems
in the preceding section. The main steps in the procedure are
outlined in Art. 144. Let,
FORCES IN RITE'S INERTIA GOVERNOR _ 395_^
W
M = — = the mass of the body which is pivoted at 0;
a) = the angular velocity of the wheel about the shaft center *S;
fl = the angular velocity of the mass M about the pivot 0;
u =aQ = the linear velocity of P relative to 0 ;
a = -T= the angular acceleration of the wheel ;
a' = — 7 = the angular acceleration of the mass M about 0 ;
7o=the moment of inertia of the mass M with respect to the
pivot 0, and
/ = the moment of inertia of the rtiass M with respect to the
• center of gravity G.
Acceleration of Any Particle P. — According to Coriolis' law
(Art. 130), the acceleration of the particle P (Fig. 4016) has three
components as follows:
(1) The acceleration that P would have considering the wheel
to be stationary and the mass to be rotating about 0; this accelera-
tion may itself be resolved into,
A radial component <tQ,^ along PO, and a tangential component
u-T. = ua' perpendicular to PO;
(2) The acceleration of that point of the wheel which is coin-
cident with P; this component may hkewise be resolved into,
A radial component poP' along P/S, and a tangential component
p-T = pa perpendicular to PS.
ai
(3) The acceleration 2uoi along OP,
These components of the acceleration of the point P are shown
in Fig. 401(6). No attempt has been made to show the magnitudes
of the accelerations by the lengths of 4^he vectors.
It will be found convenient to replace the components pw^ and
pa by four other components, parallel and perpendicular, respect-
ively, to the lines SG and PG. These four components may be
obtained by direct resolution or by considering the motion of the
wheel to be replaced by a rotation and a translation (Art. 134).
Thus, at any instant, the rotation of the wheel about its shaft, /S,
with angular velocity co and angular acceleration a may be con-
sidered as a rotation with the same angular velocity and acceleration
396 FORCE, MASS, AND ACCELERATION
about a parallel axis in the wheel which passes through the
point G, and a translation of the wheel in a direction perpendicular
to SG. The components pw^ and pa which, as found above, were
due to the rotation of the wheel about its shaft S will now be
replaced by (1) the acceleration of the point P due to the rotation
of the wheel about G and (2) the acceleration due to the translation,
which is common to all points of the wheel. The accelerations of
the point P of the wheel (that is, the point on the wheel which is
coincident with the point P of the mass Tkf), then, are.
and
xoy^ along PG
x-j' = xa perpendicular to PG,
which are due to the rotation of the wheel about G; and,
rco^ parallel to SG,
and
ra perpendicular to SG^
which are due to the (curvihnear) translation of the wheel.
Moment of Effective Forces about the Pin 0. — The moment of the
effective forces for the whole body with respect to the pin 0 will
now be found since the rotation of the mass about the pin 0 is the
cause of the movement of the valve mechanism. And, the moment
of the effective forces must be found in order to determine the
moment of the external forces which are brought into action as a
result of this rotation and which are exerted on the valve mechan-
ism. For, by D'Alembert's principle, the sum of the moments
of the external forces must be equal to the sum of the moments
of the effective forces.
The accelerating or effective force for the particle P of mass m
must have a component in the direction of each of the acceleration
components, the magnitude of each component of the force being
the product of the mass m of the particle and the acceleration
component. These seven components of the effective force for
the particle are shown in Fig. 401, acting at the particle P. The
sum of the moments of these forces, for all the particles, about the
pin 0 may be found as follows :
(1) The components maO,^ and 2muco are collinear and act
through the pin 0. The resultant of these components, therefore,
FORCES IN RITE'S INERTIA GOVERNOR 397
for the whole body, is a force which acts through 0. The sum of
the moments of these components about the point 0, then, is
equal to zero and hence, these forces have no influence in the prob-
lem under discussion. Their only effect is to produce a pin pres-
sure at 0.
(2) Consider next the forces mrco^. These forces arise from the
translation given to the body. All of these forces, then, are parallel
to SG and all have the same sense. Therefore, their resultant is
a single force (Art. 145) the magnitude of which is,
The line of action of this resultant force passes through G (Art.
145), as shown in Fig. 402. The sum of the moments of the mroa^
fr=eh=2x Area of 0 S G
Fig. 402.
forces about the pin 0 (which, of course, is equal to the moment
of their resultant), then, is (Fig. 402),
Mm^Xf.
(3) Consider next the mra forces. These forces also arise
from the translatory motion which is given to the wheel (and mass
M). All of these forces then are perpendicular to SG and have the
398 FORCE, MASS, AND ACCELERATION
same sense. Their resultant, therefore, is a single force (Art. 145)
having a magnitude equal to
l^mra = Mra.
This resultant force also passes through G. The sum of the
moments of the mra forces with respect to the pin 0, then, is
(Fig. 402),
MraXc.
(4) The moment of the mxoP' forces will be found next. These
forces are the normal forces due to the rotation of the wheel (and
mass M) about an axis through G. They pass, then, through the
axis of rotation and their resultant, therefore, is a single force
having a magnitude equal to Mxo? which acts tlirough G (Art. 22).
But, G is the mass-center of the mass M and hence, the quantity
MxcxP' is equal to zero since x, the distance from the mass-center to
the axis of rotation, is zero. Therefore, the sum of the moments
of the mxoP' forces about the pin 0 is equal to zero.
(5) The mxa forces are the tangential forces which also arise
due to the rotation of the wheel (and mass M) about the mass-
center G. The resultant of these forces is a couple having a moment
equal to la (Art. 146) in which I is the moment of inertia of the
mass M with respect to the axis through its mass-center. The
sum of the moments of the mxa forces, then, with respect to the
pin 0 (or any other moment-center in the plane of motion, Art.
15) is equal to
la.
(6) Lastly, the mca forces will be considered. These forces
are the tangential forces which arise due to the rotation of the mass
M about the pin 0. The sum of their moments, then, about 0,
the axis of rotation, is (Art. 146),
in which 7o is the moment of inertia of the mass M with respect to
the pin 0. This moment, however, for governors as constructed, is
smaU compared with the other moments since a' is small. There-
fore, the moment loa' will be neglected in the subsequent discus-
sion.
FORCES IN RITE'S INERTIA GOVERNOR 399
The Moment of the External Forces and the Governing Action. —
As noted above the moments of the external forces must be equal
to the sum of the moments of the effective forces. Or, if the
effective forces were reversed, external forces would have to act
on the body such that they would hold the reversed effective
(inertia) forces in equilibrium. Hence, external forces must act
on the mass M such that the moments of the external forces will
hold in equilibrium the moment (Fig, 402),
Tc = Mm^f=Meh(a^ (1)
of the centrifugal force, and the moment
Ti = Mrac+Ia = OL{Mrc-{-l), (2)
which is called into action only when the angular acceleration, a,
exists.
It will be noted, therefore, that if the angular velocity, w,
of the flywheel (and engine) remains constant, that is, if the
angular acceleration, a, of the flywheel is equal to zero, the
moment {MehcxP') of the centrifugal force has a constant magnitude
and must be equilibrated by the moment of the spring tension.
Further, if co is constant the only external moment acting on the
mass M is the moment of the spring tension since a, and conse-
quently Ti, is equal to zero. Thus, the moment Ti, unlike the
moment Tc, is called into existence only when there is a change in
speed of the flywheel and hence it does not require an opposing
moment except that supplied by the resistance of the valve gear
as the valve gear adjusts itself to the new speed.
To explain further the governing action in the Rite's governor,
let it be assumed that the motor is running at a constant mean
speed wo; the moment Tc{ = MehoP') of the centrifugal force is then
just balanced by the moment of the spring tension which may be
denoted by Ts. Now, let the load on the motor be reduced; the
excess of the effort over the resistance will cause the moving parts
of the motor (flywheel, shaft, etc.) to rotate with an increased
speed, coi, and the moment Tc wiU increase from Mehco^ to Mehu)i^.
The difference, Meh(oii^—i^^), is the excess of the moment of the
centrifugal force over that of the spring tension and is available
for producing a movement of the governor parts. However, if
400 FORCE, MASS, AND ACCELERATION
this difference alone is depended on for moving the valve mechan-
ism, the speed coi must be considerably in excess of the mean speed
0)0 before the governor and attached valve will move. On the
other hand, the very instant the change in speed begins, the angular
acceleration, a, of the mass M comes into existence and, with it,
the unbalanced moment Ti which is available for moving the valve
gear. In the ideal governor, therefore, there is needed a moment,
Tc, of a centrigufal force which is just sufficient to fix a mean speed
wo, and a moment, Ti, of considerable magnitude which arises from
the rate of change of the angular velocity of the flywheel and which
provides for the adjustment of the governor.
Distribution and Position of the Mass M. — An inspection of
equation (2) shows that, with a given value of r, the magnitude of
the moment Ti depends on M and /. With a heavy flywheel, a
is likely to be small; hence, to make Tt large, either M must be
made large or, for a given M, I may be made large by constructing
the swinging mass M in two parts which are removed a considerable
distance from the center of gravity, G, and are joined by a bar
(see Figs. 404 and 405).
Further, the governing action of the swinging mass will depend
largely on the relative position of the points 0, S, and G. Thus,
in Fig. 403, let a circle be drawn having a diameter equal to OS or
e and let the sense of rotation of the shaft be assumed clockwise.
As discussed above there are three moments {Mrac, la, and
Meh{oii^ — cjdQ^)) which are effective in producing the adjustment of
the governor. Now these three moments may or may not have
the same sense, depending on the location of the point G. If a
counter-clockwise sense is denoted by + and a clockwise sense
by — , the following table indicates the sense of each of the
three moments about 0 for a location of G in each of the four
regions, I, II, III, and IV indicated in Fig. 403.
I II III IV
Mrac + — — +
la + + + +
Me/i(cOl2-COo2) + - + -
For the most powerful governing action, the mass-center, (7,
of the mass M, therefore, should be located in region I. If (7 is
FORCES IN RITE'S INERTIA GOVERNOR
401
located in region III, the moment Tt becomes equal to {I — Mrc)a
and this quantity may be positive or negative according as 7
is greater or less than Mrc. With G located in regions II and IV,
that is, to the left of the vertical diameter, the moment of the cen-
trifugal force and the moment la are opposite in sense. In a gov-
ernor in which the whole mass is concentrated near to G so that
7 is small, it is permissible to locate G in region IV; with a. gov-
ernor of this type the engine may, therefore, rotate in either direc-
tion without changing the governor.
Fig. 403.
In modern governors of the Rite's type I is made very large and
hence the moment la is very large compared with the moment
Mrca. Therefore, in these governors, the mass-center of M may
be located on the circle or even in region II if it is desirable to do so
in order to decrease the centrifugal moment Mrca and, in conse-
quence, the size of the spring required.
From Fig. 403, it will be noted that the moment of the cen-
trifugal force Mroi^ is 2ilfco2 times the area of the triangle OSG)
hence, this moment is directly proportional to the area of the tri-
angle OSG. From this fact, it appears that the size of the spring
required varies directly as the area of the triangle OSG', and, to
obtain as light a spring as possible, consistent with good regula-
402
FORCE, MASS, AND ACCELERATION
tion, the triangle OSG should be made as small as possible. If G
is kept in region I, this object is accomphshed by locating G either
near to S at G', or near to 0 at G'' (Fig. 403). In the governors
shown in Fig. 404 and 405 it will be noted that G is located near to S.
ILLUSTRATIVE PROBLEM
431. In Fig. 404 is shown a Rite's inertia governor designed for a 6-h.p.
Nagle engine. Experience shows that from 0.5 to 8 ft.-lb. of energy per
Fig. 404.
h.p. should be stored in the governor weight in moving through its arc (work
done in stretching the spring). In the design, 5 ft.-lb. per h.p. was assumed.
Find the inertia weight required and the modulus of the spring from the follow-
ing data: 0/S=e =3.4 in.; A = 1.7 in.; mean speed = 250 r.p.m.; Iq = 9A in. =
unstretched length of spring; Zi = ll in. = minimum length of spring; 1-2 = 12.2
in. = maximum length of spring; p=3.8 in. = moment arm of spring tension.
Solution. — Since the moment of the centrifugal force (Tc) must equal the
moment of the spring tension {Tg) we have the equation,
That is,
Tc = Ts.
ra;2/=TmX3.8,
FORCES IN RITE'S INERTIA GOVERNOR
403
or,
W
—0}%e = TmXS.S (since rf = he),
where Tm denotes the mean value of the spring tension. In order to determine
the vahie of Tm, the work stored in the governor is equated to the work done
in stretching the spring. Thus,
12
Hence,
Tm = 300 \h.
Using this value of Tm in the above equation, we have,
From which,
w=ni\h.
The length of the spring, when the governor is in its mean position is ^(12.2 +
11.0) = 11.6. in. Hence the modulus of the spring is.
300
300
11.6-9.4 2.2
136 lb. per in.
PROBLEM
432. In the Rite's governor shown in Fig. 405, the inertia weight is 111 lb.
and the tension of the spring in mid-position is 500 lb. If e = 7 in., /i = 2.6 in.,
r = 3 in., and p =8.2 in.; find,
Fig. 405.
404 FORCE, MASS, AND ACCELERATION
(a) The normal speed of the engine, that is, the speed when the governor
is in mid-position, and
(6) The power of the governor in ft. -lb. per h.p. if the engine is rated at
50 h.p., assuming that the spring stretches from 20 in. to 22 in. when the
governor moves through its whole range.
Ans. «=268 r.p.m.; 1.67 ft.-lb. per h.p.
CHAPTER X
WORK AND ENERGY
176. Introduction. — In the preceding chapter the relations
between force, mass, and acceleration were developed from New-
ton's laws of motion and applied to the motion of bodies under the
action of unbalanced forces. As already noted, the quantities
involved directly in Newton's laws are force, mass, and acceleration.
But, acceleration involves the quantities velocity, distance, and
time. Now, in many problems in engineering, it is convenient
to use certain other quantities, the more important of which
are: work, power, energy, impulse, and momentum. The ex-
pression for each of these quantities is a combination of some of
the six quantities (force, mass, acceleration, velocity, distance, and
time) which are involved in Newton's laws of motion. Thus,
force and distance combine to measure work; force, distance, and
time combine to measure power; mass and velocity combine to
measure momentum and some forms of energy; force and time com-
bine to measure impulse, etc. Although the conceptions of these
quantities are more or less a result of our experience with physical
phenomena, the exact relations between them, as expressed in
certain principles to be developed in the following pages, are based
on the definite fundamental laws of Newton.
The present chapter is devoted to a discussion of the meaning
and use of work, of energy, and of certain principles which involve
these two quantities. Although no fundamental physical laws
other than those of Newton are used in developing the principles
of work and energy, nevertheless, the method of analysis which
makes use of work and energy, possesses certain advantages over
the method which makes use directly of force, mass, and accelera-
tion, even in certain types of problems which involve only rigid
bodies having rather simple types of motion such as translation,
rotation, and plane motion. And, in dealing with non-rigid bodies
having unordered motion, that is, motion in which the particles
405
406 WORK AND ENERGY
of the mass system (body) do not follow definite known paths,
the principles of energy are of particular importance. In fact,
the study of the behavior of non-rigid bodies in general, such
as water, steam, gas, and air is largely based on the principles of
energy and, hence, these principles play an important part in
hydrauUcs, thermodynamics, meteorology, etc.
§ 1. Work and Power
177. Work Defined. — A force does work on a body if the body
on which the force acts is moved so that the displacement of the
point of application of the force has a component in the direction
of the force. The amount of work done by a force is the product
of the force and the component of the displacement of its applica-
tion point in the direction of the force. The work done by a force
may also be expressed as the product of the component of the force
in the direction of the displacement of its application point and
the displacement. The component of the force in the direction
of the displacement of its application point is often called the
working component. And the component of the displacement in
the direction of the force is called the effective displacement.
178. Algebraic Expressions for Work Done by a Force. — The
mathematical expression for the work, w, done by a force, F, in a
displacement, s, of its application point depends on the way the
force varies during the displacement. Several important special
cases are considered here.
/. The force is constant in magnitude and in direction and
agrees in direction with the displacement as, for example, the force
exerted in lifting a body vertically upward with a uniform acceler-
ation. The amount of
work done is.
I I w = t-s.
I I
i I
'^yyyy.^^yy^yyy^yyy.;i^;!^^
II. The force is con-
jPj^ ^Qg stant in magnitude and
in direction but does not
agree in direction with the displacement (Fig. 406). The amount
of work done is,
w = F cos 6 s = Ft- s,
WORK DONE BY A FORCE
407
in which F cos d is denoted by Ft since F cos d is tangent to the path
of the point of appHcation.
///. The force varies in magnitude but not in direction and the
direction agrees with that of the
displacement as, for example, the
force exerted in compressing a
hehcal spring or the steam pres-
sure against the piston of a steam
engine after cut off. Thus, in
compressing a spring, the force
corresponding to any displace-
ment, s, is F (Fig. 407) and
this force may be assumed to
remain constant in an infinitesi-
mal displacement, ds. Hence,
according to case I, the work
done by the force F in the dis-
placement ds is dw = Fds and,
the total work done on the
spring, as F varies from its initial to its final value, is
Fig. 407.
W
<
Fds.
In order to evaluate the integral by the method of calculus, F
must be expressed in terms of s. That is, the manner in which F
varies with s must be known.
IV. The force varies in magnitude and in direction as, for
Fig. 408.
example, the pressure of the connecting rod on the crank-pin of an
engine (Fig. 408). The expression for the work done by the force
408 WORK AND ENERGY
is found by the same method as was used in case III except that
the tangential component of the force must be used. Hence,
w
This expression appHes whether the displacement is along a circular
path or not. But when the displacement takes place in a circular
path the elemental displacement ds is expressed by the equation
ds = rdd. Whence,
w
Ftds= I Ftrdd= 1 Tdd,
in which T is the torque or moment of the force about the center of
the circular path. And if the torque remains constant during an
angular displacement, 6=62 — 61, then,
Xe2
d6 =
w=T\ d6=T(62-6{) = T'6.
Thus, in one revolution, the work done by the force F having a
moment T\^w=T-2'k. And if n revolutions occur per unit of time,
then the work done per unit of time is,
w=T'2Trn.
179. Work Done by a Couple. — Since the magnitude of the
moment of a couple is the algebraic sum of the moments of the
two forces which constitute the couple, and since the moment of a
couple is the product of either force and the perpendicular distance
between the action lines of the forces, it follows from the above
discussion that the work done by a couple, having a moment T, in
a. displacement d6, is
w= TdQ,
when the moment of the couple varies. And the work done is,
w=T'6,
when the moment of the couple is constant in the displacement.
Hence, the work done by a couple having a constant moment is the
product of the moment of the couple and the angular displace-
ment of the couple.
WORK A SCALAR QUANTITY 409
180. Work a Scalar Quantity. Sign and Units of Work.—
Work is a scalar quantity. Thus the work done by one force
may be added (algebraically) to the work done by another force
regardless of the directions of the forces or of the displacements
of their points of application. And, the work done on one body
of a system may be added (algebraically) to the work done on
the other bodies of the system in order to obtain the total work
done on the system, regardless of the manner in which the bodies
move. The fact that work is a scalar quantity, frequently makes
it convenient to use this quantity under conditions which make
the use of vector quantities difficult or impossible.
It is convenient to regard the work done by a force as having
sign. Work is positive when the working component of the force
and the displacement of its application point agree in sense; and
work is negative when the working component and the displace-
ment are opposite in sense. Thus, a force which retards the motion
of a body does negative work on the body.
The unit of work is the work done by a unit force acting through
a unit distance and hence, depends on the units used for force and
distance. Thus, the more common units for work in the gravi-
tational (engineers') system of units are the foot-pound, inch-
pound, meter-kilogram, etc. No one-term names are given to the
units of work in the gravitational system of units. The common
units of work in the absolute system of units are the dyne-centi-
meter, which is called an erg, and the joule, which is IC ergs.
For large units of work the horse-power-hour and the kilowatt-hour
are used. For a definition of these units see Art. 184.
181. Graphical Representation and Calculation of Work. — In
calculating the work done by a variable force, by the calculus
method, the working component. Ft, of the force must be expressed
in terms of the displacement s. If it is impossible to express Ft
in terms of s, or if, when possible, the expression for Ft is complex
and difficult to use, as in the case of the tangential effort on the
crank-pin of a steam engine, the relation between Ft and s may be
expressed graphically by means of a graph or curve, and the work
done may be found from the graphical diagram as follows : If values
of Ft and s are plotted on a pair of rectangular axes for all positions
of the application point of the force F, the curve joining the plotted
points is called a tangential-force-space {Fts) curve (Fig. 409).
In most problems, only a sufficient number of values of Ft are
410
WORK AND ENERGY
plotted to make it possible to draw a reliable Ft-s curve, values of
Ft being plotted more frequently when the value of Ft is changing
the more rapidly. The work done by a variable force F as
shown in the preceding article is w
ds. But, Ftds repre-
sents an elemental part of the area (Fig. 409) between the Ft-s
curve and the s-axis. And, the total area under the Ft-s curve
between any two ordinates corresponding to abscissas si and
S2 is,
area =
Ftds.
Therefore, the work done by a force in any displacement s is
represented by the area under the tangential-force-space curve
between the ordinates si and S2. This diagram is called a work
diagram. In determining the
amount of work represented by
the work diagram, the scales
used in plotting the Ft-s curve
must be considered. Thus, if
ordinates are plotted to a scale
of 1 in. = 50 lb. and abscissas to
the scale of 1 in. = 5 ft., then
each square inch of area under
the Ft-s curve represents 250
ft.-lb. of work.
Since the area of a work dia-
gram equals the product of the
average ordinate and the base, the work done by a force equals the
average value of the tangential (working) component of the force
and the length of the path described by the appUcation point.
The area of the work diagram may be found by means of a
planimeter or by dividing the area into small strips and applying
Simpson's rule. Or, in some cases, less exact methods may be
employed in estimating the area.
^Displaceraent
Fig. 409.
ILLUSTRATIVE PROBLEMS
433. A helical* spring (Fig. 410) having a modulus of 200 lb. per in. is com-
pressed s = 4 in. by an axial load. How much work is done by the (variable)
load in compressing the spring?
CALCULATION OF WORK
411
Solution.— li Py denotes the force corresponding to any compression, y,
of the cpring, then from case III we have,
But,
-r
Pvdy.
Py = 200y
Hence,
= I '200y dy
200 s2 200(4)2
2 2
= 1600in.-lb.
The expression
200s-^
(when s = 4)
may be written,
Fig. 410.
w = —X - Xs = -^Xs = area of triangular work diagram
2 2
= average force times total displacement
= area of rectangular diagram having the same area as the tri-
angular diagram.
60" 90" 120"
-One half of crank circumference-
FlG. 411.
180"
434. The tangential-effort diagram for a steam engine (similar to Fig. 411)
is drawn to the following scales: 1 in. of ordinate =24 lb. per sq. in. of
piston area and 1 in. of abscissa = a 30°-arc of the crank-pin circle. The area
under the curve is found to be 11.5 sq. in. Find the work done on the crank-
pin per square inch of the piston area per stroke (one-half revolution), if the
crank length is 7.5 in. iVlso find the total work done per stroke, the diameter
of the piston being 14 in.
412
WORK AND ENERGY
Solution. — A 30°-arc of the crank-pin circle = 3.92 in.
1 sq. in. of the work diagram = 24X3.92 = 94.2 in.-lb. Work done per
stroke per square inch of piston =94.2 XI 1.5.
= 1083 in.-lb.
= 90.2 ft.-lb.
Total work per stroke = 90.2 X
7rX(14)2
= 13,900 ft.-lb.
436. In the design of punching machines (see Fig. 429) it is important to
know how much work is done in
punching a hole in a plate. Tests
show that the work-diagram for steel
is approximately of the form shown
by the heavy curved line in Fig. 412.
This diagram may be assumed, witfe-
out serious error, to be equal to the
triangular work-diagram in which the
maximum pressure P corresponds to a
shearing strength in the steel plate of
60,000 lb. per sq. in. Find the work
done in punching a |-in. hole in a f-in.
steel plate.
Fig. 412.
Solution. — Pmax = shearing area X 60,000
=7rc?<X 60,000
=7r XI X t X60,000 = 103,000 lb.
The work done in punching the hole, assuming a triangular work diagram is,
w = average pressure times thickness of plate
103,000 5
"2^8
= 32,200 in-lb.
PROBLEMS
436. A box weighing 80 lb. is pulled up an inclined plane by a force, P, of
60 lb. as shown in Fig. 413. The coefficient of friction is I. Find the work
done by each force acting on the box if it moves 20 ft. Find the total work
done on the box. Ans. 42.8 ft.-lb.
437. A rope which weighs 5 lb. per foot and which is 500 ft. long, is sus-
CALCULATION OF WORK
413
pended by one end from a drum. How many foot-ponnds of work must be
done to wind up 200 ft. of the rope?
438. Water is pumped into an elevated tank from a reservoir at the rate of
10 cu. ft. per second. How much work is done, per pound of water pumped,
in raising the level of the water in the tank 12 ft. if the initial level is 40 ft.
above the surface of the reservoir? Assume that the frictional resistance of
the water in the pipe is equivalent to an additional lift of 4 ft. Ans. 50 ft.-lb.
Fig. 413.
Fig. 414.
439. Two forces P,P (Fig, 414) exert a constant turning moment (couple)
on the hand-wheel of a large valve. The wheel is 18 in. in diameter. How
much work is done in closing the valve if 8 revolutions of the hand-wheel are
required and each force has a magnitude of 20 lb.
Fig. 415.
Fig. 416.
440. The steam indicator card (Fig. 415) is drawn to the following scales:
1 in. of ordinate = 100 lb. per square inch and 1 in. of abscissa = 5 in. of the
stroke of the piston. The area of the indicator card is found to be 2.5 sq. in.
and the length of the diagram is 3 in. (stroke = 15 in.). The diameter of the
piston is 14 in. Find the work done per stroke by the steam on the piston.
What is the average steam pressure in lb. per sq. in. (mean effective pressure)
which will do the same amount of work? Ans. w = 16,000 ft.-lb.
414 WORK AND ENERGY
441. The screw of the bracket clamp (Fig. 416) moves vertically 1 in. when
the hand-wheel is turned 4 revolutions. A helical spring having a modulus
of 200 lb. per inch is compressed by turning the hand-wheel. The average
frictional moment of the screw is 20 in. -lb. What is the average turning
moment applied to the hand-wheel in compressing the spring 3 in.?
182. Work Done on a Body by a Force System. — So far, the
work done on a body by a single force has been considered. In
general, however, a body is acted on by a force system, and, in
order to find the work done on the body in any displacement, the
work done by the whole force system must be found. In many
cases the simplest and most direct method of obtaining the work
done on the body is by finding the algebraic sum of the works done
by the forces of the system. In certain cases, however, it is con-
venient to find the work done on a body by a force system from the
resultant of the force system. In this connection the following
propositions are important:
(a) The work done by a pair of equal opposite and collinear forces in any
displacement of their application points is zero, provided that the dist nee
between the points of application of the forces remains constant. Although
the resultant of two such forces is always zero the work done, in general, is not
zero. Thus, if one end of a hehcal spring is attached to a fixed body and a
force is apphed gradually to the other end, the applied force and the reaction
of the fixed body are equal but the work done on the spring is not zero. It will
be noted, therefore, that although the resultant of the internal forces in any
body (whether rigid or not) is zero since they occur in pairs of equal, opposite,
and collinear forces (Newton's third law), the work done by the internal
forces of a body is not, in general, zero except for rigid bodies.
(b) The work done on a body by a concurrent force system is equal to
the work done by the resultant force of the system. It is convenient, as a
rule, to assume the point of application of the resultant force to be the point of
concurrence of the forces.
(c) The work done on a rigid body, in any angular displacement, 0, by a
system of couples in a plane is equal to the work done by the resultant couple,
XT. Thus, from Art. 179, the work done is
w= I
i:T-dd.
Or, if the resultant couple remains constant in the displacement, then
w = i:,T-e.
{d) The work done by the earth-pull (weight) on a body (whether rigid or
not) in any displacement equals the product of the weight of the body and the
WORK DONE BY A FORCE SYSTEM 415
vertical displacement of the mass-center of the body irrespective of its lateral
displacement. The work done by the earth-pull is positive if the center of
gravity of the body is lowered and negative if the center of gravity is raised.
It will be noted that the earth-pull on a system of particles (body) consti-
tutes a special parallel force system and that the weight of the body is the
resultant of the system.
In determining the work done by a force system which gives
a motion of translation, rotation, or plane motion to a rigid body
there are certain special cases of these motions for which it
is especially convenient to use the resultant in finding the work
done on the body. A brief discussion of these special cases
follows.
By the method of Art. 18, any coplanar system of forces may
be resolved into a system of concurrent forces acting at any chosen
point in the plane and a system of couples in the plane. The
resultant of the concurrent forces is a force (Art. 22); and the
resultant of the system of couples is a couple (Art. 27). Hence the
work done on a rigid body by the resultant of such a force system
is the sum of the works done by the resultant force and couple.
For convenience, the mass-center of the body will be chosen for the
point of concurrence of the concurrent force system for the reason
that the resultant force always agrees in direction with the accel-
eration of the mass-center (Art. 152). Further, the resultant force
is equal to the algebraic sum of the components of the forces in the
direction of the acceleration of the mass-center of the body. It
will be denoted, therefore, by '^Fq^, and the working component of
the resultant will be tangent to the path of the mass-center and
will be denoted by 2/^^ The resultant couple is the algebraic
sum of the moments of the forces about the mass-center and will be
denoted by XT. The work done on a rigid body by a coplanar force
system, therefore, is
/'S2 /"ei _
w= ( :XFafds+ I 2Tdd, (1)
Jsi Jei
in which s denotes the distance moved by the mass-center along
its path and d denotes the angle through which the body turns.
For certain important special cases of the motion of rigid bodies,
this expression reduces to simpler forms as follows.
J. Uniformly Accelerated Rectilinear Translation. — XT is zero,
2F^ is constant and acts in the direction of s. But d and s are
416
WORK AND ENERGY
the same as the a and s of every other point in the body. Hence
equation (1) reduces to
w=i:Fa'S=i:Fa-s (2)
II. Uniformly Accelerated Rotation about an Axis through the
Mass-center. — 2F^ is zero and 2T remains constant. Hence,
w=XT'd.
(3)
III. Plane Motion in which the Mass-center has Uniformly
Accelerated Rectilinear Motion and the Body has Uniformly Accel-
erated Angular Motion. — Si^^ is constant and acts in the direction of
s, and ZT is constant. Hence,
w = ZFa-'s+7:f-d. ...... (4)
ILLUSTRATIVE PROBLEM
442. A solid cylinder of radius r ft. and of weight W lb. rolls down an inclined
plane without slipping. Find the work done on the cylinder while rolling
down the plane if the plane is s ft.
long and makes an angle of <f>
degrees with the horizontal (Fig.
417).
Solution. — The forces acting on
the cylinder as shown in Fig.
417(a) are the weight W, the
normal pressure A^, and the fric-
tion F. By introducing two equal
and opposite forces, FF, at the
mass-center (Fig. 4176), and then
resolving W into x- and y-com-
ponents, the original three forces may be resolved into a force, T^ sin <^ — F,
and a couple, Fr, as shown in Fig. 417(c). From equation (4) the work done,
then, is,
w = {W sin <p-F)s-\-Fr-d.
But the displacement, s, of the mass-center equals the length, s, of the plane.
And, s = rd in which 6 is the angular displacement of the cylinder. That is,
Fig. 417.
Whence,
rd.
w = W sin <f>'S — Fs+Fs
ft.-lb.
= TT sin 0 •
Thus, it will be noted that F does no work, for, if there is no slipping, the point
of application of F moves always perpendicular to F. Likewise N does no
work since its point of application has no displacement in the direction of the
UNITS OF POWER 4l7
force. Therefore, the work done on the body is the work done by W. But,
the work done by W is Wh (Art. 182), and from the diagram it will be noted
that h= s sin </>. Hence the work done on the body is w = Ws sin <> which
agrees M^ith the result found above.
183. Power Defined. — The term power as used in mechanics
is defined as the rate of doing work. The use or function of many
machines depends upon the rate at which they do work as well as
upon the amount of work performed. Thus, some machines such
as electric generators, steam engines, etc., are rated in terms of the
power they are able to develop.
If the rate of doing work is constant, the power, P, developed
may be defined by the expression,
in which w is the work done in time t. If the rate of doing work
varies, the power' at any instant may be defined by the expression,
dt'
184. Units of Power. — Power, like work, is a scalar quantity.
The unit of power may be any unit of work per unit of time.
Thus, in the gravitational system of units, the foot-pound per
second (ft. -lb. per sec.) and kilogram-meter per second are common
units, whereas in the absolute system, the dyne-centimeter per
second (erg per sec.) or joule per second are in common use.
In many problems in engineering, however, it is more convenient
to use a larger unit of power than those mentioned above. In the
gravitational system of units these larger units are the British
or American horse-power (h.p.) and the force de cheval or Conti-
nental horse-power. They are defined as follows:
One British or American horse-power = 550 ft. -lb. per sec .
= 33,000 ft.-lb. permin.
One Continental horse-power = 75 kilogram-meters per sec.
= 4500 kilogram-meters per min.
And in the absolute system, the larger units are the watt and kilo-
watt which are defined as follows :
One watt = 10-^ ergs per sec.
One kilowatt = 1000 watts.
418 WORK AND ENERGY
The watt and kilowatt are used extensively in electrical engineering.
They may be converted into British horse-power by means of the
relations
One horse-power = 746 watts
One kilowatt =1.34 horse-power.
And, for approximate computations, it is convenient to use 1
horse-power = f kilowatt or 1 kilowatt = f horse-power.
For expressing very large quantities of work, the units used are
the horse-power-hour (h.p.-hr.) and the kilowatt-hour (kw.-hr.).
A horse-power-hour is the work done in one hour at a constant rate
of one horse-power. Thus,
One horse-power-hour =33,000X60=1,980,000 ft.-lb.
Similarly, one kilowatt-hour = 1.34X1,980,000 = 2,650,000 ft.-lb.
185. Special Equations for Power. — If a force, F, remains
constant in a given displacement of its application point and acts
in the direction of the displacement as, for example, the draw-bar
pull of a locomotive, the work done in one unit of time is Fv, in
which V is the velocity of the application point, that is, the distance
moved through in one unit of time. Hence, if F is expressed in
pounds and v in feet per second, the horse-power developed by
the force (or the body exerting the force) is,
If the velocity varies, the above equation expresses the horse-power
at the instant when the velocity is v. If the force does not act in
the direction of the displacement of its application point, the
working component may be used in the above equation. And, if
the force agrees in direction with the displacement but varies in
magnitude as, for example, the pressure of the steam against the
piston of a steam engine or the tangential effort against the crank-
pin, then the value of F (or Ft) at any instant may be used to obtain
the power at that instant. However, the average power during a
given cycle (or many cycles) of operations is generally more useful
than the instantaneous power. Thus, in the case of a steam
engine the average horse-power is expressed by,
_2Plan
^"33000'
SPECIAL EQUATIONS FOR POWER
419
in which P is the mean effective pressure (lb. per sq. in.), a is the
piston area (sq. in.), I is the length of stroke (ft.), and n is the
number of revolutions per minute (r.p.m.). For, Pa is the average
force (lb.) which acts through a distance l-2n (ft.) per minute, the
number of strokes per minute being 2n in a double-acting engine,
and hence the work (ft.-lb.) done per minute (power) is Pa'l-2n
and the horse-power is as given above.
If a couple having a constant moment, T, acts through a given
angular displacement of 6 radians, the work done is 7"- ^ (Art. 179).
And, if the couple turns through co radians per unit of time, the
work done per unit of time is To). But, w is the angular velocity
of turning. Hence, if the moment of the couple is expressed in
pound-feet and w in radians per second, the horse-power developed
by the couple is
If the angular velocity at which the couple is turning is not
constant, the above equation expresses the horse-power at the
instant at which the velocity is oo. But, in most cases the average
horse-power during a given cycle of operations is of more use than
the instantaneous value.
PROBLEMS
443. A locomotive exerts a constant draw-bar pull of 35,000 lb. while
increasing the speed of a train from 30 to 45 miles per hour. What horse-
power does the engine develop (a) at the beginning of the period; (6) at the
end of the period? What is the average
horse-power during the period?
444. A man in turning the crank on the
winch of a crane was found to exert the
forces shown in Fig, 418 at the positions indi-
cated. Plot (freehand) carefully a tangen-
tial-effort diagram, the crank (radius) being
15 in. long. Estimate from the diagram the
mean tangential effort and calculate the mean
horse-power developed by the man assuming
that he turns the crank at a constant speed of
40 r.p.m. Ans. 0.55 h.p. Fig. 418.
445. Two pulleys are keyed to the same shaft 10 ft. apart. One pulley is
driven by a belt from an engine. The other pulley is belted to, and drives a
machine. If the first (driving) pulley receives 3 h.p. from its belt what torque
420 WORK AND ENERGY
is transmitted to the shaft (and driven pulley), assuming that the shaft
rotates at a constant speed of 150 r.p.m. .4ns. 105 Ib.-ft.
446. What indicated horse-power will the engine referred to in Prob. 440
develop if it operates at a constant speed of 250 r.p.m. and is double acting?
447. A generator develops 500 k.w. and delivers 450 k.w. to a machine
shop. A price of 4 Qents per kw.-hr. is paid. Does the machine shop pay for
power or for work? What is the cost to the machine shop per day of 8 hours?
448. A certain machine requires 5 h.p. for its operation. If the machine
is in use 6 hr. per day how many foot-pounds of work are deUvered to the
machine in one day?
449. An accumulator loaded to a pressure of 750 lb. per square inch is used
in emergencies to supply the lubricant to a step bearing of a vertical steam
turbine. If the ram is 10 in. in diameter and has a stroke of 15 ft., what horse-
power does it deliver if one hour is required to complete the stroke?
§ 2. Energy
186. Energy Defined. — The energy of a body is the capacity
of the body for doing work. Work may be considered to be done
by forces, as in the preceding section, or, since forces are exerted by
bodies, work may also be considered to be done by the bodies
which exert the forces, the work being done by virtue of the
energy which the bodies possess. A body may have the capacity
to do work (possess energy) due to a variety of conditions or states
of the body. Thus, energy may be classified as mechanical energy,
heat or thermal energy, chemical energy, electrical energy, etc.,
depending on the state or condition of the body by virtue of which
it is capable of doing work. Our knowledge of all the conditions
which render bodies capable of doing work is far from complete,
but experience shows that any of the forms of energy may, under
the proper conditions, be transformed into any of the other forms.
Mechanical energy is of particular importance in connection
with the kinetics of bodies and is therefore considered at some
length in the following pages. The other forms of energy are dis-
cussed briefly in Art. 191. Mechanical energy is divided into
potential energy, or energy of position or configuration, and kinetic
energy or energy of motion.
From the definition of energy it follows that energy, like work,
is a scalar quantity. Thus, the energy of any mass-system is the
algebraic sum of the energies of the various particles of the system
regardless of the directions of motion of the particles.
POTENTIAL ENERGY 421
The units of energy are the same as the units of work discussed
in the preceding section.
187. Potential Energy. — The potential energy of a body is the
capacity of the body for doing work due to its position or configura-
tion. Thus, the water above a mill dam posesses potential energy
which may be used in driving a water wheel; a compressed
spring and the compressed steam in a boiler are capable of doing
work by virtue of the relative positions (configuration) of their
particles.
The potential energy of a body may be defined quantitatively
as the amount of work which a body is capable of doing against
forces, in passing from the given position or configuration to some
standard position or configuration, assuming that no other change
in the state or condition of the body takes place. This definition,
however, does not lead to a definite quantity for the potential
energy of the body for a given configuration, unless the work done
by the body (mass-system) depends only on the initial and final
configuration of the mass-system and not at all on the paths
described by the parts of the system while coming to the standard
state. Mass-systems for which this condition is fulfilled are called
conservative mass-systems, and the force system which acts on
such a mass-system while its potential state changes is called a con-
servative force system.
The potential energy of conservative systems, only, will be
considered herein since, in most kinetics problems which involve
non-conservative systems the kinetic energy of the system k
of greater importance in the solution of the problems. The
most common case of a non-conservative system is that in which
the mass-system does work against frictional forces, such as
sliding and journal friction and the friction of the particles devel-
oped in deforming an inelastic body. Conservative mass-systems
occur frequently in engineering problems. In fact any rigid
body under the action of a force system in which friction does not
occur (or may be considered negligible) is a conservative system,
provided, of course, that no change in the state or condition of the
body except that of configuration takes place. A common example
of a conservative system is that of the earth and an elevated body
(whether rigid or not). The work done on the body in any dis-
placement is equal to the earth-pull (weight) of the body times the
vertical displacement of the center of gravity of the body (Art.
422 WORK AND ENERGY
182), regardless of the intermediate positions occupied by the body
in moving from one position to another position. Another example
is that of an elastic body, for, if the body is elastic the energy
possessed by the body when in a given strained condition, that is,
for a given configuration of its particles, is the same regardless of
the relative displacements of the particles which occurred while
being put in the given strained condition. The standard configu-
ration may be arbitrarily chosen, but, for convenience, it is so
chosen that the potential energy of the body is positive. Thus, in
the case of the earth and an elevated body the earth is considered
fixed and the standard configuration occurs when the body is in
contact with the earth. A discussion of the mathematical test for
a conservative system is beyond the scope of this book.
188. Kinetic Energy. — The kinetic energy of a body is its
capacity for doing work due to its motion. Thus, by virtue of its
kinetic energy, a body is capable of doing work against forces which
change its motion. For example, a jet of water does work on a
tangential water-wheel; a steam forging hammer does work on the
material which is deformed by the hammer; the rotating flywheel
on a punching machine does work in punching the hole in the
metal plate, etc.
The kinetic energy of a body at any instant may be defined
quantitatively as the amount of work that the body is capable of
doing against forces which destroy its motion, that is, which bring
it to a state of rest. The expression for the kinetic energy of a
body (mass-system) should, therefore, contain a quantity (velocity)
which is a measure of the motion of the body and also a quantity
(mass, moment of inertia, etc.) which is a measure of the (kinetic)
property of the body that has an influence in governing its change
of motion. The '' velocity of a mass-system," however, is, in
general, an indefinite and meaningless phrase since the velocities
of the various parts of a system are not the same. Hence, an
expression for the kinetic energy of a particle is first obtained and,
since energy is a scalar quantity, the kinetic energy of a system
of particles (mass-system) is the algebraic sum of the kinetic
energies of the particles. However, the expression for the kinetic
energy of a particle is of considerable importance in itself since in
many problems a physical body may be regarded as a particle
without introducing serious errors.
189. Kinetic Energy of a Particle. — In Fig. 419, let P be a par-
KINETIC ENERGY OF A PARTICLE
423
tide of mass m in a body (assumed rigid for convenience only)
which moves so that P travels from position P' to P" , along the
path shown, while its velocity decreases from v at P' to zero at P"
due to the forces against which the particle does work. The work
done on the particle by the forces (which form a concurrent system)
is equal to the work done by their resultant R. Or, w
■■ ( Rtds.
Fig. 419.
But, by definition, the kinetic energy, Ek, of the particle is the work
which the particle does against the forces. Hence, the defining
equation for the kinetic energy of a particle is,
Et^—w
= - I Rtds.
This expression may be transformed so that Et is expressed in terms
of m and v by means of the following relations.
Thus,
Rt = mat
E,
dv , ds
-y
= — I mat(is=— I m—i
Js, Js^ dt
Rtds
matds =
424 WORK AND ENERGY
Therefore, the kinetic energy of a particle of mass m having a
velocity v is equal to \rmP. That is,
PROBLEMS
450. By making use of the equations of Art. 124, prove that the kinetic
energy of a particle having uniformly accelerated rectilinear motion is ^ mv^.
451. The German long-range gun which shelled Paris from a distance of
approximately 76 miles was 118 ft. long. The muzzle velocity of the pro-
jectile was not far from 5000 ft. per sec. The diameter of the projectile was
8.15 in. and its weight was 264 lb. It attained a height of about 24 miles,
was in flight about 3 min., and reached Paris with a velocity of about 2300 ft.
per sec. (For a description of the gun see Journal A. S. M. E., Feb., 1920.)
Neglecting the energy due to the rotation of the projectile, calculate the kinetic
energy of the projectile as it left the gun, and also its energy at the end of
its flight. Find the loss of kinetic energy per second during the flight.
190. Kinetic Energy of a Body. — Since energy is a scalar
quantity the kinetic energy of a body (v^^hether rigid or not) is
the algebraic sum of the kinetic energies of its particles. Hence,
for any mass-system,
It is convenient, however, to express the kinetic energy of a
rigid body in terms of the mass (or some other kinetic property
such as moment of inertia) of the whole body and the velocity of
some particular point in the body (as, for example, the mass-
center) or the angular velocity of the whole body. Thus, for
rigid bodies having the special motions of translation, rotation,
and plane motion the expressions for the kinetic energy are found
as follows:
1. Translation of a Rigid Body. — All parts of the body have the
same velocity at any instant whether the motion is rectilinear
translation or curvilinear translation (Art. 132), that is, v is con-
stant. Hence,
But, 2m is the mass of the body and may be denoted by M,
Therefore,
KINETIC ENERGY OF A. BODY
425
II. Rotation of a Rigid Body. — The angular velocities of all
particles are the same at any instant, that is, the angular
velocity co of any particle is the angular velocity of the body.
The linear velocity, v, of any particle, P, of mass m, at a distance,
r, from the axis of rotation, 0, (Fig. 420) is equal to roi. Hence,
the kinetic energy of the body is
Et = Simt;2 = i2m(a)r)2 = Jco^Smr^.
But, Smr^ is the moment of inertia of the body with respect to the
axis from which r is measured, that is, the axis of rotation, and
is denoted by Iq. Thus, Jo = Zmr^.
Therefore,
Fig. 420.
Fig. 421.
///. Plane Motion of a Rigid Body. — The angular velocities of
all particles are the same at any instant, that is, the angular
velocity co of any particle is the angular velocity of the body.
Further, as shown in Art. 134, the motion of the body at any
instant may be considered to be a combination of a rotation about
an axis through any point, 0, in the plane of motion and a trans-
lation defined by the motion of 0. Hence, the velocity, v, of any
particle P of mass m (Fig. 421) is the resultant of the velocity, cor,
which P is given by the rotation about 0, and the velocity, vot
which is given to all particles by the translation. And, since the
body is rigid the velocity cor has a direction perpendicular to r.
Thus,
tP={o3r)^-\-vo^-\-2voo)r cos 6.
In Fig. 421, let 0 be the origin and, for simplicity, let the rc-axis
have the same direction as vq. The kinetic energy of the body may
426 WORK AND ENERGY
then be found in terms of the mass of the whole body, the angular
velocity of the body, and the linear velocity of one point (in this
case the point 0) in the body as follows :
Ei = Zimi^ = ii:miZ^'^-\-vo^+2voo:r cos d)
= |2mcoV+j2mz;o^+2mi'ot*^?' cos d
= ^o)^Xmr^-\-ivo^^ni-ho)Vo'^mr cos 6.
But, Smr^ is the moment of inertia of the body with respect to the
axis through 0 from which r is measured. Thus, Smr^ = 7o- Fur-
ther, r cos d = y, whence, 2mr cos 6 = Zmy = My in which M is the
mass of the body. Therefore,
J^t = hIoc^^+iMvo^+Myc^vo.
Now since the point 0 is any point in the plane of motion it may be
chosen at the mass-center. That is, the motion of the body may
be resolved into a rotation about an axis through the mass-center
and a translation defined by the motion of the mass-center. If the
point 0 is made the mass-center, then, ^ = 0, Jo becomes I, and
vo becomes D. Hence, the kinetic energy is given by the expression,
Et = iI(o^+iMv^. ........ (2)
It is important to note that, although plane motion of a rigid
body may be resolved at any instant, into a rotation about an axis
through any point in the plane of motion and a translation defined
by the motion of that point, it does not follow that the kinetic
energy of the body at the
given instant is the kinetic
energy due to the rotation
plus the kinetic energy due
to the translation, unless
the assumed rotation is
about an axis through the
mass-center of the body
and the translation is de-
fined by the motion of the
mass-center.
Alternative Method. — Plane motion of a rigid body may be
considered to be pure rotation about the instantaneous axis of
Fig. 422.
KINETIC ENERGY OF A BODY 427
rotation (Art. 135). The kinetic energy of a body, as for example,
the connecting rod represented in Fig. 422, therefore, is,
in which U is the moment of inertia of the body with respect to the
instantaneous axis. But, by the parallel axis theorem (Art. 102),
Ii = l-\-Mf^. Therefore,
PROBLEMS
452. A body B weighing 60 lb. is attached by a flexible string to a spring S
(Fig. 423). The pulley over which the string passes is weightless and friction-
less. The spring has a modulus of 40 lb. per inch.
If a force P of 50 lb. is gradually applied to B, how
much is the potential energy of the spring changed? Of
the spring and body B considered as one system?
453. A slender rod, similar to the spoke of a fly-
wheel, is 3 ft. long and rotates about an axis through
one end at a constant speed of 120 r.p.m. The rod
weighs 50 lb. Find the kinetic energy which the rod
possesses. Ans. ^a; = 368 ft. -lb.
454. Two spherical bodies each weighing 20 lb. are
connected by a slender rod and revolve at 90 r.p.m. in
a horizontal plane about a vertical axis located midway Yig. 423.
between the two bodies. The center of each ball is
10 in. from the axis. The diameter of each ball is 4 in. The weight of the
rod is 6 lb. Find the kinetic energy of the system. Ans. Ek = '^0.5 ft.-lb.
455. The winding drum of a mine hoist is 14 ft. in diameter, its radius of
gyration is 6 ft., and its weight is 7 tons. A cage weighing 6 tons is raised by it.
When the cage is rising at the rate of 40 ft. per sec. what is the kinetic energy
of the system? Ans. -&* = 554,000 ft.-lb.
456. The cast iron flywheel (Fig. 424) is used on a punching machine in the
forge shop at the University of Illinois. If the flywheel is rotating at a speed
of 225 r.p.m. when the punch starts to punch a hole and has its speed reduced
20 per cent in punching the hole, how much energy does the flywheel give up;
(o) neglecting hub and spoke, (6) including hub and spokes, assuming the spokes
to have a constant mean cross-section and to run from the hub to the rim?
457. Find the kinetic energy of the rocker arm AD (Fig. 425) from the
following data: OB = 10 in.; 0^=3 ft.; angular velocity of crank 05 = 50
r.p.m.; 0 = 30°; length of rod AD =4.5 ft.; and the weight of the rod is 22.7 lb.
428
WORK AND ENERGY
468. Find the kinetic energy of the connecting rod, the motion of which is
described in Prob. 386.
Fig. 424.
191. Non-mechanical Energy. — Experience shows that some
bodies are capable of doing work (possess energy) by virtue of
certain states or conditions of their parts, the nature of which is not
definitely enough known to make it possible to determine then-
energy by the methods used in the preceding articles. Energy
which cannot be determined directly as potential or kinetic energy
is called non-mechanical or non-dynamic energy. Thus, heat or
thermal energy, chemical energy, and electrical energy are forms
of non-mechanical energy.
A body is capable of doing work by reason of its heated state or
condition since by giving up its heat it may do work, under
favorable conditions, as in the case of steam in the cylinder of a
steam engine. Energy possessed by a body by virtue of its heated
state is called heat or thermal energy.
Certain bodies are capable of doing work by reason of their
chemical state or condition. Thus, hydrogen and oxygen, under
favorable conditions, combine and give evidence of considerable
energy by transferring heat to surrounding bodies. Likewise,
carbon (coal) and oxygen combine and produce heat which as noted
above may in turn do work. Energy possessed by bodies due to
the state of their chemical elements is called chemical energy.
NON-MECHANICAL ENERGY 429
Some bodies are capable of doing work by virtue of their elec-
trical state or condition. Thus, a copper wire on an armature
moving in a field of force may develop electric current which in
turn may do work in driving a motor. Or, a charged storage
battery may do work as its electrical condition changes, etc.
Energy which arises out of the electrical conditions of bodies is
called electrical energy.
Any one of these, so-called, special forms of energy may be
converted, under favorable conditions, into mechanical energy,
and there is considerable evidence to indicate that all energy
is mechanical energy. Thus, according to this view, the heat
energy of a body could be determined as kinetic energy if the
motions of the individual particles were known. And, certain
forms of chemical and electrical energy could be determined as
potential energy if the molecular forces were definitely known.
Therefore, the energy possessed by bodies by virtue of special
states of their molecular structure are considered as non-mechanical
forms of energy, not because these special forms are necessarily
different from mechanical energy, but because the energy cannot
be determined directly as mechanical energy, and, therefore, has
to be transformed into mechanical energy and then measured.
Thus, one unit of heat energy, the British thermal unit (B.t.u.),
has a definite mechanical equivalent which carefully made experi-
ments have shown to be
1 B.t.u. = 778 ft.-lb.
However, the lack of knowledge of the molecular structure and
conditions by virtue of which bodies possess energy does not pre-
vent the appUcation of certain principles of energy to such condi-
tions. In fact, the outstanding feature concerning energy is that
certain general principles of energy, such as principles of the con-
servation of energy and of degradation of energy, etc., are the basis
upon which our knowledge of the behavior of non-rigid bodies, in
general, is built and thus they furnish a method of approach to
problems for which the principles of force, mass, and acceleration
are inadequate. They are of special importance, therefore, in the
study of hydraulics, thermodynamics, electrodynamics, physical
chemistry, etc.
In the following section certain principles concerning mechan-
ical energy are developed and applied to the motion of bodies
430
WORK AND ENERGY
(mainly rigid) in which the motion of the particles are definitely
known. And even though the principles of force, mass, and acceler-
ation may be used for many of the problems considered, never-
theless, it will be noted that even for rigid bodies the principles of
work and energy are of great importance in many kinetics problems
as met in engineering practice.
§ 3. Principles of Work and Energy
192. Preliminary. — As stated in Art. 137, in order to deal
with the main problem in kinetics, a relation is found between
the forces acting on the body, the kinetic properties (mass, moment
of inertia, etc.) of the body by virtue of which it influences its own
motion, and the change of motion (involving acceleration, velocity,
distance, etc.) of the body. This may be done by determining the
relation between the work done by the forces acting on the body
and the kinetic energy of the body, since work and kinetic energy
involve quantities in terms of which the three factors in the
kinetics problem as mentioned above are expressed. And since,
in general, the motion of all particles of a body are not the same,
the principle of work and kinetic energy will be developed for a
particle first and then extended to the motion of a body.
However, as already noted, in many problems the body may be
regarded as a particle without introducing serious errors.
193. Principle of Work and Kinetic Energy. — I. For a Par-
tide.— Iji Fig. 426 let A'
and A'' be two positions
of a body, the motion of
which changes due to the
unbalanced forces {F\, F2,
Fs, and F4) which act on
it. The body is assumed,
for convenience, to be
composed of small cubes of
different materials rigidly
attached (glued together),
j^ each cube being regarded
Fig. 426. 9^ a particle of the body.
Let P be one of the parti-
cles which describes the path shown in the figure as the particle
moves from P' to P'' while its velocity changes from vi to V2 due
lO-
\"
^^
H-
\
Oi
571
\
k
-P"
PRINCIPLE OF WORK AND KINETIC ENERGY 481
to the unbalanced (concurrent) forces which act on it. Let the
resultant of the forces acting on P be denoted by R and its com-
ponent tangent to the path of P be denoted by Rt. It will be
noted that some of the particles (cubes) have their velocities
changed (are accelerated) by forces exerted only by other particles
of the body (internal forces) whereas other particles are acted on
both by internal and by external forces.
The work done by the forces acting on the particle as it moves
^i^"
along its path from P to P \^ w= \ Rtds but, as in Art. 189,
Rtds may be expressed in terms of the mass, m, and velocity, v,
oi the particle by means of the relations,
^ dv . ds
Rt — mat, (k = -Tp and ^ = "j:-
The work done on the particle = I Rtds
r
icle= I
= I matds= I
=/
dv ,
m—rds
dt
m-rdv
vi dt
■■ I mvdv
= ^mv2^ — \mv^
= change in kinetic energy of the
particle.
Therefore, the work done hy the forces acting on a particle of a
body (whether rigid or not) during any displacement is equal to the
change in the kinetic energy of the particle in the same displacement.
Or, expressed in the form of an equation,
w = AEt = imv2^ — imvi^.
11. For a System of Particles. — The principle of work and kinetic
energy for a system of particles (body) may now be derived.
Since work and energy are scalar quantities, the work done on a
body is the algebraic sum of the works done on all the particles.
And, the change in the kinetic energy of the body is the algebraic
sum of the changes in the kinetic energies of all the particles.
432 WORK AND ENERGY
But, the work done by all the forces acting on all the particles
equals the work done by the external forces which act on the body
plus the work done by the internal forces of the body. Thus, by
writing the above equation for each particle and adding both sides
of the equations, the resulting equation is,
We-{-Wi = J2mz;2^ — ^'Lmvi^ = AEt.
That is, the work done on a system of particles {whether rigid or not)
by all of the external and internal forces in any displacement of the
system is equal to the change in the kinetic energy of the system in
the same displacement.
III. For a Rigid Body. — The principle may now be expressed
for the special case of a rigid body, for, as pointed out in Art. 182,
the internal forces in any mass-system occur in pairs of equal
opposite and coUinear forces whether the body is rigid or not, but
the work done by these forces is zero, in general, only if the appli-
cation points of each pair of forces remain a fixed distance apart,
which is the case in a rigid body. Hence, for a rigid body, Wt = 0.
Therefore,
We = AEi.
That is, the work done hy the external forces acting on a rigid body
in any displacement is equal to the change in the kinetic energy of
the body in the same displacement.
Although an absolutely rigid body does not exist in nature, the
work done by the external forces in causing the relative displace-
ments of the part-c.es in physical bodies is usually negligible in
comparison with the work done by the forces in causing the
displacement of the body as a whole.
194. Application of Principle to Special Cases of Motion of
Rigid Bodies. — By making use of the expressions developed in
Art. 182 and 190, the principle of work and kinetic energy may be
expressed for important special cases of motion of rigid bodies as
follows :
/. For a Rigid Body Having Uniformly Accelerated Rectilinear
Translation. —
XF,'S = iM{v2^-vi^).
II. For a Rigid Body Rotating about an Axis through the Mass-
center with Constant Angular Acceleration. —
2:r-6l = j7(co22-coi2).
PRINCIPLE OF WORK AND KINETIC ENERGY 433
III. For a Rigid Body Having Plane Motion Such that the
Angular Acceleration is Constant and the Mass-center has Uniformly
Accelerated Rectilinear Motion. —
2:Fa-s+i:T-d=iM{v2^-vi^)+ilic^2^-o^i^).
As noted in Art. 182 the work done by the forces acting on any
body may be found by taking the algebraic sum of the works done
by all the forces. And in more general cases of motion than those
stated above, it will be, as a rule, more convenient to find the work
done in this way rather than from the resultant of the forces.
ILLUSTRATIVE PROBLEMS
469. An engine capable of exerting a maximum draw-bar pull of 51,000 lb.
is used on a certain railroad having small grades to draw freight trains
having a maximum weight of 2000
tons, the average weight of a freight — . 51,00^ •
car with its cargo being about 45
tons. The train resistance per ton
of weight varies with the car weight
and with the speed. If an average
value of 8 lb. per ton is used and
the engine pulls a 2000-ton train
while going up a | per cent grade, how far will the train travel while its
velocity is increasing from 15 to 30 mi. per hour? How long will it take?
Solution. — The forces acting on the train are shown in the free-body dia-
gram (Fig. 427). The angle 6 (exaggerated in the diagram) for a ^ per cent
grade is so small that tan d may be considered to be equal to sin 6. Hence,
1
sin d=——.
200
We = AEic,
2000 tons •'N
Fig. 427.
XFa-S = hM{V2'-V,^),
(«■
2000X2000\ ^1 2000X2000/— 2 —^]
0OO-l<i''^----^~^)s = '^--"--^f^{u^-W)
Since,
we have.
Whence,
15,000s = 62,200X1452,
s = 6020 ft.
6020 = 2?±Mx«.
t = 182.5 sec.
=3.04 min.
434
WORK AND ENERGY
460. In the friction-board type of drop hammer used in the production of
drop forgings, the ram (Fig. 428) is attached to the lower end of a friction board
which is Hfted by friction driving rolls and these in turn are driven by spur
gears. The roUs are pressed against the board by revolving the eccentric
bearings and they may be released quickly to let the ram drop. The following
data apply to one particular hammer :
Friction board
Friction rolls
Weight of ram, Tf = 1000 lb.
Angular velocity of rolls, co = 130 r.p.m.
Diameter of rolls, d = ll in.
Normal pressure against friction board,
AT = 6000 lb.
Coefficient of friction, n = 0.3.
Total height to which ram is raised, h =
4.5 ft.
Find the distance the ram travels (a) while
being accelerated, that is, while its velocity is
being brought up to the peripheral speed of
the friction rolls; (h) while the ram is coming
to rest after the rolls are released; (c) while the
ram is moving with constant velocity, after
the acceleration ceases and before the roUs
are released. Also find the time (seconds) required to travel each of
these distances and the time to complete a cycle (total time). Assume that
the friction rolls do not slip on the friction board. Although there is some
slipping just as the rolls are pressed against the board, this assumption will not
introduce serious errors in the problem.
Solution. — The friction force, F', during the acceleration period is,
F' = fxN = 0.3 X 6000 = 1800 lb.
The peripheral speed of the rolls is,
130X27r^5.5 «orf. /
v = o)r = — — X ^2- = 6.25 ft. /sec.
During the acceleration period, the forces acting on the ram are F' and W, as
shown in Fig. 428. Applying the principle of work and energy, we have,
We = AEt,
(F'-W)si=W(v'-0),
(1800 - 1000)si = 2 ^ X (^•2^)*-
Hence,
800si=606.
si= 0.758 ft.
PRINCIPLE OF WORK AND KINETIC ENERGY
435
Aft«r the rolls are released, the ram is acted upon only by the earth-p«ll—
(weight) while it gives up its kinetic energy. Thus,
We=AEt,
1 1000
1000X83= ^If^X (6.25)2.
S3 =0.606 ft.
S2 = /l-(Si+S3)
=4.5 -(0.758 +0.606)
= 3.14 ft.
Since the distance traveled in each period is the average speed times the time,
we have.
Therefore,
Hence,
Therefore,
Likewise,
Whence,
Also,
whence.
Further,
Hence,
0.758 = ^X^1.
<i =0.243 sec. = duration of acceleration period.
0.606 = ^ X<3.
^3 = 0. 194 sec. = duration of decelerating period.
3.14 = 6.25X^2,
fe = 0.503 sec. = duration of constant velocity period.
-Vf=/-^>
= 0.528 sec. =time of falling.
= 1.468 sec. =time of complete cycle.
461. The punching and shearing machine shown in Fig. 429 has a capacity
for punching a 2^-in. hole in a ^-in. steel plate. The pinion shaft (and fly-
wheel) is driven at 220 r.p.m. from a counter shaft by means of a belt drive to
the tight pulley on the pinion shaft. Each operation of punching a hole
(punching cycle) causes a fluctuaton (decrease) in the speed of the flywheel.
The " coefficient of speed fluctuation " is 0.8, that is, the speed of the flywheel
decreases 20 per cent in each punching cycle. If the work done in punching
the hole is all supplied by the flywheel, what moment of inertia should the
flywheel have? What is the moment of inertia of the flywheel on the machine
436
WORK AND ENERGY
as actually designed (as shown in Fig. 429), neglecting the material in the hub
and spokes? If the difference in the belt tensions, T2 — T1, is 286 lb. and the
pinion shaft turns through an angle of 230° while the hole is being punched,
how much work is done on the shaft (and flywheel) by the belt while the hole is
being punched? The diameter of the pulley is 22 in.
Solution. — The maximum value of the force P required to punch the hole.
Punch attachment
[P-« Force exerted by
punch
J) 1
Pinion gear
Pinion shaft
-f
Fig. 429.
using 60,000 lb. per square inch for the ultimate shearing strength of the
material of the plate, is
P=7rdX<X60,000
=7rXlX^X60,000
= 235,500 lb.
The work done in punching the hole, assuming a triangular work diagram
(see Prob. 435), is,
P, 1 235,500 ^cya^n- ^u
^y = o X o = — -A =58,875 m.-lb.
2 2 4
= 4900 ft.-lb. which is supplied by the flywheel.
For the flywheel we have then,
4900 = ^7(a;22-coi2)
= .7[(-^^)^-(0.8X-^)]=95.57.
7=51.3 slug-ft.2
Whence,
PRINCIPLE OF WORK AND KINETIC ENERGY
437
Hence, the moment of inertia of the flywheel should be 51.3 slug-ft.^ in
order that the speed be decreased not more than 20 per cent. Assuming the
flywheel to be made of ciast-iron which weighs 450 lb. per cubic foot, the
moment of inertia of the flywheel as actually designed, neglecting hub and
spokes, is
=\-u^-mrv-i-M-^mr]
1
XI. 36X0. 375X450X6. 64
2 32.2
= 74.3 slug-ft.2.
The work done on the flywheel while the hole is being punched is,
w = 'LT'd
= 286x{4x230Xj|o
= 1050ft.-lb.
462. A solid homogeneous cylinder rolls up an inclined plane without
slipping (Fig. 430). The weight
of the cylinder is 120 lb. and
its diameter is 3 ft. The angle,
0, of inclination of the plane is
15°. If the velocity of the center
of the cylinder is 20 ft. per sec.
just as it comes in contact with
the incline, how far up the plane
will the cylinder roll?
Solution. — The forces acting
on the cylinder while rolling up
the plane are shown in Fig.
430. And, the work done by
the forces, as was shown in Prob. 442, is W sin </>
Thus,
Fig. 430.
Or,
Also
Hence,
-W sin 0-s = K^y2'+/'«^2')-K^^i'+/<oi2).
TF sin </>•« = \{Mvi^ -\-l lai^) , since Th and o>2 are zero.
v = roi = 1.^03.
1 120
120X0.2588X8 = =-^^^ {20y+\i^K^Xl .b
Whence,
2 32.2
746J-372
31.1
2V2 32.2
= 36 ft.
1(3
438
WORK AND ENERGY
PROBLEMS
463. A body weighing 80 lb. is projected along a rough horizontal plane
with a velocity of 8 ft. per sec. It comes to rest in a distance of 10 ft. Find
the coefficient of kinetic friction. Ans. /i = 0.099.
464. An automobile which weighs TT lb. is moving at the rate of 30 miles
per hour when it comes to the foot of a hill. Power is then shut off. The
slope of the hill is 1 ft. in 50 ft. How far will the machine coast up the hill
if the total frictional resistance (parallel to the road) is 0.08 W^
Ans. s = 301 ft.
465. A sphere rotates about an axis through its center. Its speed is
increased from 600 rad. per min. to 90 rad. per sec. while it turns through 10
revolutions. If the moment of inertia of the sphere with respect to the axis
of rotation is 20 lb. sec.^ ft. (or slug-ft.^), find the moment of the couple acting
on the sphere.
466. The winding drum of a mine hoist is 15 ft. in diameter, its radius of
gyration is 6 ft., and its weight is 8 tons. A cage weighing 6 tons is raised by it.
If the cage is rising at the rate of 60 ft. per sec, at what level should the power
be shut off in order that the cage may stop at the surface without braking?
Neglect friction. Ans. 103.5 ft.
467. The hand-operated screw
press shown in Fig. 431 is used for
embossing, lettering dies, punching
thin plates, etc. The diameter of
the screw is 2| in. The screw has
triple threads with a pitch of 2^ in.
Each ball weighs 100 lb. and the
diameter of each ball is 9 in. If
the balls are revolved at 60 r.p.m.,
what is the maximum size (diame-
ter) of hole that can be punched in
a j-in. steel plate, assuming the
shearing strength of the steel to be
60,000 lb. per square inch and the
efficiency of the screw to be 15 per
cent. Also assume the work dia-
punching the hole to be triangular. (See Probs. 435 and 461.)
Ans. d = O.QOm.
Fig. 431,
gram for
468. A shearing machine has 3 h.p. delivered to it by the belt. Every two
seconds an operation occurs which requires seven-eighths of all the energy
supplied during the two seconds; the other one-eighth of the energy is
required to overcome the friction of the machine. During each operation
the speed of the flywheel decreases from 120 to 80 r.p.m. Assuming that
the work done in shearing is done by the flywheel, what should be the
CONSERVATION OF ENERGY
439
moment of inertia of the flywheel?
what is its radius of gyration?
If the weight of the flywheel is 400 lb.
469. A solid cylinder weighing 500 lb. is rolled up an inclined plane by
means of a descending weight B (Fig. 432). The diameter of the cylinder is
4 ft. The pulley over which the rope runs is assumed to be frictionless and
weightless. Body B weighs 300 lb. If the cylinder starts from rest at the
bottom of the incline, with what velocity will its center reach the top?
470. A solid disc 18 in. in diameter is mounted on a shaft 4 in. in diameter
(Fig. 433). The shaft rolls, without slipping, on two inclined tracks. The
disc and shaft weigh 120 lb. The angle of inclination, <^, is 15° and the incline
Fig. 432.
Fig. 433.
is 8 ft. long. If the disc starts from rest at the top, what will be the velocity
of its center at the bottom of the incline? Ans. y = 3.46 ft./sec.
195. Conservation of Energy. — One of the greatest achieve-
ments of the nineteenth century was the recognition and statement
of the principle of the conservation of energy. Like Newton's
laws of motion it is an inductive generalization from observation of,
and experience with, physical phenomena. The principle states
that in any change of the state or condition of an isolated material
system the total amount of energy of the system remains constant.
By an isolated system is meant one on which no bodies external to
the system have any effect on the system. Hence, an isolated
system neither gives nor receives energy. Thus, the distribution
of energy within the isolated system may be altered and the
various forms of energy changed into other forms but the total
amount of energy remains constant. Or, as sometimes stated,
energy may be transformed or transferred but cannot be created
or destroyed. As noted in Art. 191, the principle of 'conservation
of energy is of particular importance in the study of material
systems which possess non-mechanical energy, although it is also
of much value in the study of mechanical energy. Although an
isolated system does not exist in nature, certain systems approach
440 WORK AND ENERGY
closely thereto, as for example, the earth and a falling body, pro-
vided that the action (and reaction) between the earth and body is
large compared with the resistance of the air and the attractions
of other bodies on the falling body. Again, although external
forces act on a system of bodies, the work done by the forces may
be zero (or negligible) as in the case of a simple swinging pendu-
lum and the earth in which the pull of the string on the bob is al-
ways normal to the displacement of its application point and the
effect of the air is negligible.
196. Relation between Potential and Kinetic Energy for Con-
servative Systems.— For non-rigid mass-systems it was found
(Art. 193) that,
We-{-Wi = AEi,
but Wi, which represents the work done by the internal forces
in any displacement of the system, is equal (but opposite in
sign) to the change in potential energy of the mass-system for
the given displacement,* that is,
Wi=—AEp.
Therefore,
We = AEt-{-AEp.
Now if the work done by the external forces which act on a mass-
system is zero, then, We = 0 and hence,
AEt+AEp = 0.
Or,
AEt=-AEp.
* This fact may be shown as follows: By definition the change of potential
energy of a body is the work which the body does against forces in passing
from one configuration to another configuration, providing no other change in
the state of the body (such as a change in velocity) takes place. Then from the
equation,
We-\-Wi=AEt;,
we obtain,
or,
Wi= —We,
since AEt is zero if no change of the velocity state occurs. But, by definition
— We is the potential energy of the system in its initial configuration. As
explained in Art. 187, i^c is a definite quantity only when the forces are con-
servative.
POTENTIAL AND KINETIC ENERGY 441
That is, in any displacement of a conservative mass-system Xor~
one on which the external forces do no work), the gain (or loss)
in the kinetic energy of the system is equal to the loss (or gain)
in the potential energy of the system.
If the system is not conservative, the transformation of the
energy within the system from potential to kinetic energy (or vice
versa) is accompanied by an increase in other forms of energy
such as heat energy. And, although the total energy remains con-
stant, the mechanical (potential and kinetic) energy does not
remain constant. The relations between potential and kinetic
energy for conservative systems, only, are considered herein.
ILLUSTRATIVE PROBLEMS
471. A body which weighs 50 lb. falls from a height, h, of 50 in. upon a
helical spring (Fig. 434), the modulus of which is 800 lb. per inch. What is
the maximum shortening, s, of the spring?
Solution. — Little error will be introduced by assuming the spring and body
to be perfectly elastic; by neglecting the air resistance on the falling body;
and by neglecting the inertia of the spring. If this is done, the earth, the body,
and the spring, may be considered to be a conservative system in which the
earth is assumed to be fixed. Thus the mechanical energy of the system re-
mains constant and hence,
But AJ5Ja; = 0, since the kinetic energy of the system is zero
at the beginning and at the end of the displacement. And
the change in the potential energy of the system is,
Or,
in which h and s are expressed in inches. This equation
states that the loss of potential energy of the body is
equal to the gain of potential energy of the spring. Solving
the equation, we have,
50(50+s)=400s2
Or,
8s2-s-50=0.
Whence,
s= 2.56 in. Fig 434.
472. A simple pendulum (Fig. 435) consists of a cord having a length, r,
of 4 ft. and a bob (assumed to be a particle) having a weight, W, of 6 lb. The
442
WORK AND ENERGY
pendulum is displaced an angle, d, of 60°. What will be the velocity, v, of
the bob when in its lowest position, if air resistance is neglected?
Solution. — The tension T in the string does no work. Thus, the earth
and bob may be treated as a conservative system. Hence,
^E^
Wh-
6(4-4 cos 60°)
whence,
1 6
2 32.2 '
y = 11.36 ft./sec.
473. Solve Prob. 462 by use of the
principle expressed by the equation,
LEic+AEp = Q.
Solution. — Ab shown in Prob. 442,
the only force which does work is the
earth-pull. Hence, the earth and the cylinder form a conservative system
Thus,
^Ep=-^E1c,
Ws sin 0 = ilMv'^ + i/a;2) .
This is the same equation that was obtained and solved in Prob. 462.
Substituting the given values in the above equation as was done in Prob.
462, the value of s is found to be 36 ft.
Note. — ^All three of the above problems may be solved by the principle of
work and kinetic energy. It is important, however, to note the relation
between potential and kinetic energy for conservative systems.
§ 4. Efficiency. Dissipation of Energy
197. Efficiency Defined. — The efficiency of a machine, as for
example, a steam engine, an electric motor, a chain hoist, a jack
screw, etc., is the ratio of the energy output of the machine in a
given period of time to the energy input in the same period, pro-
vided that no energy is stored in the machine which becomes avail-
able at a later period. By input is meant the amount of energy
received by the machine, a portion of which is transformed or
transmitted into the work or energy for which the machine is
designed. The work done or energy delivered by the machine is
called the output. Thus, denoting efficiency by e, we have.
e =
energy output
energy input
or e
power output
power input *
DISSIPATION OF ENERGY 443
As already noted, in the transformation and transference of
energy (which is the main function of many machines), some of the
energy always takes the form of a lower grade of energy (heat
energy) and thereby becomes unavailable for the particular process
for which the machine is used. The amount of energy which thus
miscarries or leaks out in the process is spoken of by various names,
such as lost energy (or lost work), energy leak, dissipated energy,
etc. Since dissipation of energy occurs with every physical
process, the output is always less than the input and, therefore,
the efficiency is always less than unity.
The efficiency as defined above is the over-all efficiency
of the machine. Certain parts of the machine, however, may
have their individual efficiencies; and the over-all efficiency
is the product of the efficiencies of the several elements of the
machine.
198. Dissipation of Energy. — The work done against friction
is the most frequent cause of dissipation of energy in machines.
Energy dissipated in doing work against frictional forces is trans-
formed into heat energy. Electrical resistance in connection with
electrical machinery also causes a loss of available energy by devel-
oping heat. The work done against friction is, in some machines, a
necessary evil to be reduced to a minimum as in the case of prime
movers, bearings, teeth of gears, etc. ; whereas, in other machines or
machine elements, the main object of the machine is to dissipate
all the energy received by the machine, as in the case of friction
brakes and absorption dynamometers.
Two typical examples involving dissipation of energy in
machines are here considered: (1) a bearing in which the loss of
energy is made as small as practicable by lubrication and (2) a
simple Prony brake or absorption dynamometer, the object of
which is not only to absorb all the energy delivered to it but also
to measure that energy (or power). Another class of dynamom-
eters measures the power delivered by a machine without absorbing
(dissipating) the energy. Dynamometers of this class are called
transmission dynamometers.
199. Work Lost Due to Friction in Bearings. — The dissipa-
tion of energy in bearings of machines of various kinds reduces the
useful energy transmitted or transformed by the machine and is,
therefore, an important factor in the design and operation of such
machines. For example, in textile machinery, which involves a
444
WORK AND ENERGY
large number of small parts moving at relatively high velocities, a
large part of the load is due to friction.
Various types of bearings are used to meet the different needs.
Thus, for shafts subjected to a considerable end thrust, as, for
example, a screw-propellor shaft or the shaft of a centrifugal pump,
a collar thrust bearing having several collars may be used, whereas
for slow speeds, such as occur in certain types of rotary cranes,
the end thrust due to the load and weights of the moving parts
is usually carried by an ordinary flat pivot or step bearing.
Fig. 436 represents an ordinary flat pivot in which the work
(and power) lost in friction is to be found. Certain assumptions
must be made. It will be assumed that
the coefficient of friction for the rubbing
surfaces is constant over the whole area
of contact and that the normal wear at
any point (wear normal to the rubbing
surfaces) is proportional to the work of
friction. Thus, according to this as-
sumption, the pressure after wear has
occurred will not be uniform but will be
a maximum at the center where no wear
occurs and will decrease towards the
edge, since the relative velocity of the
rubbing surfaces is zero at the center
and increases towards the edge. It will,
therefore, be well to cut out the material near the center of the
bearing to avoid crushing of the material. Let p be the intensity
of pressure on any elementary area 2irpdp, at a distance p from
the axis of the shaft. The total pressure P, then, is
P = 27r j ppdp.
And from the assumption made above, the normal wear, n, is
Fig. 436.
in which fc is a constant.
n = kpp or p =
Therefore,
Jn k
27rn
= ^r(^2-ri).
kp'
(1)
WORK LOST DUE TO FRICTION IN BEARING 445
The frictional force at any distance p from the axis is,
= fxp2Trpdp
2Trn,
And the frictional moment Tf for the whole area is,
= ^(r2^-ri^) (2)
By eliminating n and k from (1) and (2), the following equation is
obtained,
Tr=^(r^+n) (3)
And, if CO is the angular velocity of the shaft in radians per second,
the work lost per second, %, is,
i^,= r,a; = ^(r2+ri) .... (4)
If P is expressed m pounds and ri and r2 in feet, then Wf is ex-
pressed in foot-pounds per second and the horse-power lost in
friction is,
., . tiPoi r2±ri
(h.p.)/=^X-^-. (5)
In order to increase the amount of rubbing surface and thereby
diminish the intensity of pressure, a collar bearing with two or
more collars may be used. Equations (8), (4), and (5) also apply
to such bearings. It will be observed that the coefficient of fric-
tion is an important factor in the problem of reducing the dissi-
pation of energy due to friction and this fact suggests the great
importance of the choice of bearing materials and of lubricants.
Various other types of bearings may be treated in a manner
similar to that used above.
446
WORK AND ENERGY
200. A Simple Dynamometer. Prony Brake. — A simple dyna-
mometer, commonly called a Prony brake, is shown in Fig. 437.
^ is a flanged pulley keyed to a rotating shaft. The power trans-
mitted by the shaft is absorbed and measured by the brake.
B, B are bearing blocks against which the pulley develops a fric-
tional resistance, the magnitude of which is varied by adjusting
the nuts C, C. D is the frame or beam with its end E resting on
the platform of a weighing scale. When the pulley is running,
the beam develops an additional pressure on the platform, due
to the friction of the pulley on the bearing blocks. The work lost
in friction and the power developed by the shaft may be found as
follows :
It will be assumed that the scales are adjusted to read zero
when the beam rests on the platform and the pulley is not running.
Thus, the scale reading is a measure of the pressure, P, at E due to
the friction developed on the bearing blocks when the pulley is
running. Since the brake frame is in equilibrium under the action
of the forces, N, P, and F (F denotes the total frictional force on
the two blocks), the sum of the moments of P and F about the
axis of the shaft must equal zero. Or,
Fr = Pa.
And the work done by the frictional moment Fr in one second is,
Wj = Fro3
= Pact),
in which co is expressed in radians per second. And, since co =
27rn
A SIMPLE DYNAMOMETER 447
where n is the number of revolutions per minute (r.p.m.) of the
shaft and pulley, the expression for Wf becomes,
2TPan
irPan
30
If P is expressed in pounds and a in feet, then Wf will be expressed
in foot-pounds and the horse-power developed by the frictional
moment (and hence by the shaft) is,
, _ TPan
^"30X550
_ TPan
~ 16,500 •
This expression may be simplified if the dynamometer is constructed
so that a has a special value. It should be remembered that the
scale reading should not be used for the value of P unless the scales
are adjusted to read zero when the pulley is not running.
PROBLEMS
474. What effort, P, is required to raise a load, Q, of 200 lb. by means of
the differential chain hoist shown in Fig. 438 if e = 30 per cent, r2 = 8 in., and
ri=4 in.? Ans. P = 167 1b.
475. In a test of a jackscrew (see Fig. 169) with a screw L5 in. in diameter
and a pitch of ^ in,, it was found that a pull of 72 lb. at the end of a 15-in. lever
was required to raise a load of 2400 lb. when no lubricant was used and a pull
of 63 lb. when a hard oil lubricant was used. What is the efficiency of the
jack for each case? Ans. e = lL8%; e = 13.5 per cent.
476. The band brake described in Prob. 165 allows a certain load to lower
at a constant speed such that the brake sheaves rotate at 120 r.p.m., in
a counter-clockwise direction. What horse-power is absorbed by the
brake?
477. The dynamometer shown in Fig. 439 was devised by Lord Kelvin in
connection with the laying of the Atlantic cable for braking the cable drum as
the cable was laid out. If it is used in the test of a steam engine, find the
horse-power absorbed, using the following data : W = 400 lb. ; speed of engine =
1,50 r.p.m.; r = 4ft.; reading of spring balance is 85 lb. Ans. 36 h.p.
448
WORK AND ENERGY
478. Derive an expression for the work done by friction in the spherical
pivot shown in Fig. 179, assuming that normal wear is proportional to the
Fig. 439.
work of friction and that the coefficient of friction is the same for all parts of
the rubbing surfaces.
Ans. Work per revolution = 27rjuWr
sin^ a
a+sin a cos a.
CHAPTER XI
IMPULSE AND MOMENTUM
201. Preliminary. — In Art. 176, the statement was made that
impulse is a quantity which involves force and time, and that
momentum is a quantity which involves mass and velocity.
And the fact was noted that the use of these quantities in the
analysis of the motion of bodies requires no fundamental laws in
addition to Newton's laws of motion. However, methods which
make use of impulse and momentum offer advantages, in certain
types of problems, over the methods of work and energy (Chap.
X) and of force and acceleration (Chap. IX).
In determining the effect of forces on the motion of bodies,
thus far, by the methods of force and acceleration and of work and
energy, it has been assumed that the forces have acted on rigid
bodies during a definite (comparatively large) interval of time,
and when the forces were not constant the manner in which they
varied during the period was assumed to be known. To such con-
ditions the methods of impulse and momentum also apply, and
in many cases offer a simpler method of solution than the methods
previously discussed. Forces sometimes act, however, for a very
short (indefinite) interval of time during which neither the value
of the force at any instant nor its law of variation is known.
These forces may, nevertheless, produce very appreciable changes
in the motion (velocity) of the body. Such forces are called
impulsive forces. The bodies upon which impulsive forces act
deform under the excessive pressures produced and hence, in
determining the motions of bodies under the influence of impulsive
forces, the bodies cannot always be assumed to be rigid, as in
the preceding chapters, without introducing appreciable errors.
As examples of impulsive forces the following may be mentioned :
the force exerted on a projectile due to the explosion of the pow-
der; the action of one billiard ball on another; the force exerted
by the ram of a pile driver on the pile; the pressure between two
449
450 IMPULSE AND MOMENTUM
railway cars when making a flying coupling; the action of a steam
jet on the blades of a high-speed steam turbine; the pressure
exerted by the water in a pipe line on a valve which is closed sud-
denly. The principles of impulse and momentum are of special
value when considering the motion of bodies under the action of
impulsive forces.
The purpose of the present chapter is to make clear the con-
ception or meaning of impulse and of momentum, to develop
certain principles which express relations between these quan-
tities, and to apply these principles to problems in kinetics.
§ 1. Impulse
202. Impulse and Impact Defined. Units. — The impulse of a
constant force is defined as the product of the magnitude of the
force and the time interval during which the force acts. Thus,
if Q denotes the impulse of a force F, the impulse of the force is
defined by the equation,
Q = FAt,
provided that the force remains constant during the time interval
A^. If the force varies in magnitude but not in direction, the
impulse for an indefinitely short period of time dt, is F-dt and, for
a time interval At = t2 — ti, the impulse is.
'-£
Q= I Fdt
In order to evaluate this integral by the method of calculus, F
must be expressed in terms of t. The impulse of a force which
acts during a very short (indefinite) interval of time (impulsive
force) is also given by the above expression but, as noted in the
preceding article, the impulsive force cannot be expressed in terms
of t, since its law of variation is not known and hence, the impulse
cannot be determined directly but is found in terms of the change
of momentum as is discussed in the subsequent pages.
The impulse of an impulsive force is sometimes called an
impact, that is, an impact is a sudden impulse. The term impact,
however, is also frequently used as descriptive of the act of collision
of bodies. In some problems it is convenient to estimate the
time interval of the impulsive force and to express the impact as
LINEAR IMPULSE OF A FORCE 451
the product of an average value of the force and an assumed or
estimated time interval At = t2 — ti. Thus,
= j Vd^ = F,
Q= j Fdt = FavAt,
in which Fav denotes the average value of the impulsive force which
is assumed to act during the time interval A^. The impulse of a
force frequently is called linear impulse in contrast with the
moment of the impulse which is called angular impulse. (See
Art. 204.)
Units. — The unit of an impulse is a combination of a unit of
force and of time and hence, is a compound unit. The unit of
impulse has no special name. In the gravitational or engineer's
system of units, if the pound is selected as the unit of force and the
second for the unit of time, then the unit of impulse is the pound-
second (Ib.-sec.) or, if the kilogram is selected for the unit of force
and the second for the unit of time, the unit of impulse is the kilo-
gram-second (kg.-sec).
203. Components of Linear Impulse. — The impulse of a
force, like the force itself, is a directed or vector quantity, the
sense and action line of the impulse being the same as that of
the force. An impulse of a force, therefore, may be resolved
into components and may have a moment with respect to a
point or a line. The component, in any direction, of the impulse
of a constant force is the product of the component of the force in
the given direction and the time interval At during which the force
acts. That is,
Qx = Fj:'At; Qy = Fy'At, etc.
And, if the force varies in magnitude during the time interval
At = t2 — tij the component of the impulse is.
Qx= \ F^dt; Qy= I Fydt, etc.
Linear Impulse of a Force System. — The linear impulse, in any
direction, of a force system is the algebraic sum of the components,
in the given direction, of the impulses of the forces of the system.
452 IMPULSE AND MOMENTUM
Thus, for a force system in which the forces are constant, the
linear impulse, Qx, of the force system in any direction x is,
Or, if the forces of the system vary during the interval, then.
'-^r
204. Moment of Impulse. Angular Impulse. — The moment
of the impulse of a constant force about any point or axis is the
product of the moment, T, of the force about the given point or
axis, and the time interval, A^, during which the force acts.
Thus, by letting Lq denote the moment of the impulse of a con-
stant force with respect to an axis 0, the moment of the impulse
of the force is defined by the equation,
And, if the force varies in magnitude during the time interval,
M = t2 — ti, the moment of the impulse is expressed by the equation.
Lo
=jyodt,
in which To is the moment of the force with respect to the axis 0.
The moment of the impulse of a force is also called the angular
impulse of the force.
Angular Impulse of a Force System. — The angular impulse of
a force system about any axis is the algebraic sum of the angular
impulses of the forces of the system about the given axis. Thus,
if the forces of the system are constant, the angular impulse, Lq,
of the system about the axis 0 is,
Lo = ^ToAt
And, if the forces vary, the angular impulse of the force system
about the axis O, for the time interval At = t2 — ti, is expressed by
the equation,
Lo = JlCTodt
MOMENTUM OF A PARTICLE DEFINED 453
It should be noted that the linear (and angular) impulse of a
constant force, or of a variable force for which the law of variation
is known, is not as important a conception in the analysis of the
motion of a body as is the impulse of an impulsive force. However,
as already noted, even for constant forces which act during a
comparatively long time interval, the methods of impulse and
momentum may possess advantages over other methods.
PROBLEMS
479. A body weighing 40 lb. slides down an inclined plane in 4 sec.
The plane makes an angle of 60° with the horizontal. If the coefficient of
friction is 0.2 find the component of the linear impulse of the force system
acting on the body parallel to the plane.
Ans. Qa; = 1231b.-sec.
480. The linear impulse of the total pressure of the steam on the piston
of a steam engine is 4200 lb .-sec. The diameter of the cylinder is 14 in. and
the engine runs at 200 r.p.m. Find the average (time average) pressure (in
lb. per sq. in.) of the steam against the piston during one stroke.
481. A train having a weight of 2000 tons travels up a | per cent grade.
The draw-bar pull of the engine is 50,000 lb. and the train resistance is 8 lb.
per ton of weight. If it takes 2 min. to travel up the grade, find the linear
impulse of the force system acting on the train for the 2-min. interval.
Ans. 1,680,000 Ib.-sec.
482. A constant frictional moment of 200 lb .-ft. is applied to a rotating
drum by means of a band brake (see Fig. 185). If the moment decreases
the angular velocity of the drum uniformly from 90 r.p.m. to 10 r.p.m. while
the drum makes 30 revolutions, find the angular impulse of the band brake
on the drum. Ans. 7200 lb. -ft .-sec.
§ 2. Momentum
205. Momentum of a Particle Defined. Units. — The
momentum of a particle of a moving body, at any instant, is
defined as the product of the mass of the particle and its velocity
at the instant. Momentum, hke velocity, is a directed or vector
quantity. Furthermore, like force, momentum is represented by a
localized vector, that is, it has a definite position line. Thus, the
direction of the momentum of a particle is the same as that of the
velocity of the particle and its position hne passes through the
particle. Thus, if U denotes the momentum of a particle of mass
m and velocity v, the magnitude of the momentum of the particle
is defined by the equation,
U=mv,
454
IMPULSE AND MOMENTUM
The momentum of a particle frequently is called linear
momentum in contrast with the moment of momentum of the
particle which is called angular momentum.
Units. — The unit of momentum is a combination of a unit of
mass and a unit of time and hence, is a compound unit. The unit
of momentum has no special name. In the gravitational or engi-
neer's system of units, the unit of mass is a derived unit as explained
in Art. 141; derived from the units of force, length, and time and
hence, the unit of momentum is also a derived unit. Thus, if
the pound, foot, and second are chosen for the units of force,
length, and time, respectively, the unit of momentum is expressed
by,
1 unit of mass X 1 unit of velocity =
lib. XI sec.2_ 1 ft.
1 ft.
= 1 Ib.-sec.
X
1 sec.
It will be observed, therefore, that momentum is expressed in the
same fundamental units as is impulse.
206. Components of Momentum. Moment of Momentum. —
Since the momentum of a particle is a vector quantity it may be
resolved into components and, like any localized vector, it has a
moment with respect to any
point, the moment being defined
as the product of the magni-
tude of the momentum and the
perpendicular distance from the
position line to the point or
moment-center. If, then, Ux
and Uy denote the components
of the momentum of a particle
of mass m and of velocity v
(Fig. 440) and, if ^o denotes
the moment of momentum of
the particle with respect to the point 0, the components of the
momentum of the particle and the moment of momentum of
the particle are expressed by the equations,
Fig. 440.
and
Ux={mv)x = mvx', Uy = {mv)y = mvy,
Ho = mv-r; or, Ho = 7nVx-y — niVy-x.
LINEAR MOMENTUM OF A BODY
455
The latter expression for Hq expresses the fact that the moment of
the momentum of a particle equals the algebraic sum of the
moments of the components of the momentum of the particle.
207. Linear Momentum of a Body. — The linear momentum, in
any direction, of any body (mass-system) is the algebraic sum of
the components in the given direction of the momentums of the
particles of the body. In gen-
eral the momentums of the
particles of a body are not the
same either in magnitude or
in direction. It will now be
shown, however, that the
linear momentum of the whole
body may be found from the
mass of the whole body and
the velocity of the mass-center
of the body. Thus, in Fig.
441 are shown three particles
m', m", and m'" of a system
assumed, for convenience only,
to move in a plane. At a
given instant the particles and their mass-center are in the posi-
tions shown. The direction of the momentum of each particle is
tangent to the path of the particle and the velocity, v, of the mass-
center is tangent to the path which the mass-center describes.
The component, in the a;-direction, of the linear momentum of the
mass-system is,
C7x = mV+m'V+m''W"+
= S(mz;i).
But, ^{mvx) may be shown to be equal to Vx^m or Mvx as follows:
The x-coordinate, x, of the mass-center is defined by the principle
of moments (Art. 80) by the equation,
m'x'^-m"x"-{-m'"x"'+ . . . =x^m = Mx,
By differentiating both sides of the equation, the following equa-
tion is obtained,
.ax . f/UX . ...d.T/ ,
n^dx
456
IMPULSE AND MOMENTUM
dx'
But, -T- is the a:-component of the velocity of m' and hence,
dx' .
Similarly,
dt-'''"
dx" „ ,
dt ='^' "*"-
and
dx -
dt='-
Thus, the above equation becomes,
Similarly,
Whence,
m'v:
m'i\
MVx', Uy = MVy; and U=Mu.
That is, the linear momentum of any moving mass-system is the
product of the mass of the whole system and the velocity of the mass-
center of the system and its direction agrees with that of the velocity
of the mass-center.
208. Angular Momentum of a Rotating Rigid Body. — Let
Fig. 442 represent a rigid body rotat-
ing about a fixed axis through 0 with
an angular velocity co. The linear
velocity of any particle of the body
at a distance r from the axis of rota-
tion is ro) (Art. 115) and the linear
momentum of the particle of mass m
is mv or mrcj. Therefore, the moment
of momentum for one particle about
the axis of rotation is,
Fig. 442.
mv-r = mrco • r = mr^'cc,
and the algebraic sum of the moments of the momentums of all the
particles about the same axis is,
whence.
Ho = Smr^co = oj^mr^y
Ho = /oco.
ANGULAR MOMENTUM OF RIGID BODY
457
in which Iq is the moment of inertia of the body about the axis of
rotation.
Therefore, the angular momentum of a rotating rigid body
about the axis of rotation is equal to the product of the moment of
inertia of the body with respect to the axis of rotation and the
angular velocity of the body.
209. Angular Momentum of a Rigid Body Having Plane
Motion. — Since a plane motion of a rigid body may be considered
as a combination of a rota-
tion and a translation (Art.
134), the velocity of any
particle P (Fig. 443) is the
resultant of a velocity rco
due to the rotation of the
body with angular velocity
CO about an axis 0 and, the
velocity vq due to the trans-
lation which gives to each
particle the velocity vo of
the point 0. Therefore, the
components of the momen-
tum of the particle P are
mrco and mvo as shown in Fig. 443. And, if m{vo)x and m{vo)j,
denote the x- and ^/-components of the momentum mvo, then the
angular momentum of the particle with respect to the axis 0 is,
mro3-r-\-m{vo)x'y — m(vo)v'X,
and the angular momentum, ^o, for the whole body with respect
to the axis 0 is.
Fig. 443.
Xm(vo)vOi^
o)X7nr^-\-{vo)x'2rny—{vo)yXmx
Io<d+{vo)xMy-{vo)yMx. . .
(1)
Now, as already noted, a plane motion of a rigid body may be
considered as a rotation about an axis through any point in the
plane of motion and a translation, etc. Hence, if the point 0 is
taken as the mass-center of the body, then ^o becomes H and the
above expression reduces to
5=7(0, (2)
458 IMPULSE AND MOMENTUM
since y and x are then equal to zero and 7o becomes /. Therefore,
in general, the angular momentum of a rigid body having plane
motion, with respect to an axis in the body, is not equal to the
product of the moment of inertia of the body with respect to the
given axis and the angular velocity of the body. But, for an axis
passing through the mass-center of the body, the arigular mo-
mentum is equal to the product of the moment of inertia of the
body with respect to the axis through the mass-center and the
angular velocity of the body.
It may be noted also that if the axis about which the angular
momentum of the body is taken is the instantaneous axis of rota-
tion, equation (1) becomes Ho = Ioo} since {vo)x and {vo)y are equal
to zero. That is, the angular momentum of a rigid body having
plane motion, about the instantaneous axis of rotation, is equal
to the product of the moment of inertia of the body with respect
to the instantaneous axis of rotation and the angular velocity of
the body.
PROBLEMS
483. A small body (particle) weighing 8 lb. is attached to one end of a
string and is made to revolve as a conical pendulum (see Fig. 324). If the
body revolves at a distance of 15 in. from the axis with an angular velocity of
90 r.p.m., find (a) the linear momentum of the body, (6) the angular momentum
of the body with respect to the axis of rotation.
Ans. (a) C/ = 2.93 1b.-sec.; (6) ^ = 3.66 Ib.-ft.-sec.
484. A slender rod rotates in a horizontal plane about a vertical axis
through one end of the rod. If the rod is 3 ft. long, weighs 4 lb. per foot, and
rotates at 120 r.p.m., find its angular momentum about the axis of rotation.
Ans. fl^o = 14.05 Ib.-ft.-sec.
485. Find the angular momentum, about the axis of rotation, of the fly-
wheel shown on the punching machine in Fig. 429, when the flywheel is rotat-
ing at 225 r.p.m. Use the value of / found in Prob. 461.
486. Find the angular momentum, about the axis of rotation, of the two
spheres shown in Fig. 431, when the spheres are revolving at 120 r.p.m. Each
sphere weighs 100 lb. and is 9 in. in diameter. Ans. Ho = 1250 lb .-ft .-sec.
487. A solid cylinder weighing 128.8 lb. rolls, without slipping, down an
inclined plane. The linear velocity of the mass-center of the cylinder at a
given instant is 30 ft. per sec. The diameter of the cylinder is 18 in. Find
the angular momentum of the cylinder (a) about an axis through the mass-
center perpendicular to the plane of motion, (b) about the instantaneous axis
of rotation. (Ans.) (a) ^ = 45 lb. -ft. -sec; (&) i/i = 135 Ib.-ft.-sec.
LINEAR IMPULSE AND LINEAR MOMENTUM 459
488. The connecting rod shown in Fig. 422 is 30 in. long. It has a
constant cross-section and weighs 52 lb. The crank length OB is 5 in. and
(i>i is 300 r.p.m. The distance from i to 5 is found to be 34.68 in. and f is
found to be 23.75 in. Find the angular momentum of the connecting rod (a)
about an axis through the mass-center and (6) about the instantaneous axis
of rotation,
§ 3. Principles of Impulse and Momentum
210. Preliminary. — In order to determine the effect of a force
system on th'^ motion of a body, that is, in order to treat the usual
problem in kinetics, by means of the quantities impulse and
momentum, the relations which exist between the impulse of
the force sj^stem and the momentum of the body on which the
force system acts must be established. These relations are
expressed by means of two principles, namely, (1) the principle
of Unear impulse and linear momentum and (2) the principle of
angular impulse and angular momentum.
211. Principle of Linear Impulse and Linear Momentum. —
It was stated in Art. 152 that the algebraic sum of the com-
ponents in a given direction of the external forces acting on any
body (whether rigid or not) is equal to the mass of the body times
the component of the acceleration of the mass-center of the body
in the given direction. That is, if x denotes any direction,
^F, = Ma. (1)
Now if the forces acting on the body remain constant, then ax
will be constant, that is, the mass-center of the body will be uni-
formly accelerated. Hence, according to Art. 124,
in which vj' and vj are the a;-components of the velocity of the
mass-center of the body at the end and beginning, respectively, of
the time interval M. Therefore,
^F,'Lt=^MvJ'-Mv,' (2)
Now the left-hand member of this equation is the a;-component of
the linear impulse of the external force system which acts on the
body (Art. 203) and the right-hand member is the change in the
460 IMPULSE AND MOMENTUM
momentum of the body in the a:-direction (Art. 207). The above
equation is a mathematical statement of the principle of linear
impulse and Unear momentum for any body when acted upon by
constant forces. The principle applies, however, to the motion of
any body under the action of any force system. Thus, in general,
0'x = -if and hence, equation (1) may be written,
at
or, since M is constant,
S/^. = J(Mi;.) (3)
And, by integrating this equation, the following equation which
expresses the principle of linear impulse and linear momentum is
obtained.
Jh Jvx'
Jh JVx
Or,
Fjli^Mv/'-Mv,', (4)
■■c
in which, as before, Vx' and Vx are the a:-components of the velocity
of the mass-center of the body at the end and at the beginning,
respectively, of the time interval t2 — ti. The principle of linear
impulse and linear momentum then, as expressed in equations (2)
and (4), may be stated in words as follows: The algebraic sum of
the components, in any direction, of the impulses of the external forces
abating on a body during any time interval is equxil to the change in the
linear momentum of the body in the same direction during the same
interval of time, or, stated in the form of an equation,
Qx = AUx,
in which x represents any direction. The particular forms in
which Qx and Ux are expressed depend upon the type of force
system, the type of motion, and the kind of body (see Art. 214
for expressions which apply to rigid bodies having particular
types of motion).
It should be noted also that equation (3) expresses an impor-
tant principle which may be stated in words as follows: The alge-
braic sum of the components, in any direction, of the external forces
ANGULAR IMPULSE AND ANGULAR MOMENTUM 461
acting on any body is equal to the rate of change of the linear mo-
mentum of the body in the same direction.
212. Principle of Angular Impulse and Angular Momentum. —
It was shown in Art. 146 that if a rigid body rotates about a fixed
axis, the algebraic sum of the moments of the external forces
about the axis of rotation is equal to the product of the moment
of inertia of the body about the axis of rotation and the angular
acceleration of the body. That is,
^To = Ioa (1)
Now, if the forces acting on the body remain constant, then a
will be constant, that is, the body will rotate with uniformly accel-
erated motion and hence, from Art. 125,
A^
Therefore,
Sro-A^ = /oO)2-/oa)i. ...... (2)
Now the left-hand member of this equation is the moment of the
impulse (angular impulse), about the axis of rotation, of the force
system acting on the body, and the right-hand member is the
change in the moment of momentum (angular momentum) of the
body with respect to the axis of rotation.
If the forces acting on the body are not constant, then a will
not be constant, its value at any instant being — (Art. 120).
Hence, in general,
27-0 = /o^,
and since /o is constant, this equation may be written in the form,
SJo = j^(/o<o) (3)
Now by integrating this equation, the following equation, which
expresses the principle of angular impulse and angular momentum
for a rigid body rotating about a fixed axis, is obtained.
) '^Todt= \
d{IocA}).
'S I T^di = /oo)2 — /o<oi (4)
462 IMPULSE AND MOMENTUM
The principle, however, appUes to any mass-system having any
type of motion. The principle may be stated in words as follows:
The algebraic sum of the angular impulses, about any axis, of the
external forces acting on any body (mass-system) for any period of time
is equal to the change in the angular momentum of the body about the
same axis during the same interval of time. Or, expressed in terms
of the symbols which already have been defined, the principle is
expressed by the equation,
Lo = AHo.
The particular forms in which Lo and Ho are expressed depend on
the type of force system, the type of motion, and the kind of
body. (See Art. 214 for expressions which apply to a rigid body
having particular types of motion under the action of constant
forces.)
It should be noted also that equation (3) expresses an impor-
tant principle which may be stated in words as follows: The alge-
braic sum of the moments of the forces acti?ig on a rotating rigid body
about the axis of rotation is equal to the rate of change of the angular
momentum of the body about the same axis. The principle, however,
is not restricted to the motion of a rigid body rotating about a
fixed axis. It applies to any body having any type of motion. It
should be noted, however, that, in general, the angular mo-
mentum of a body is not equal to 7co.
213. Method of Analysis of the Motion of a Body by Means
of Impulse and Momentum. — It was noted in Art. 144 that, in
the analysis of the motion of any body under the action of an
unbalanced force system, relations must be found which involve
(1) the forces acting on the body, (2) the kinetic properties of the
body, and (3) the kinematic properties of the motion of the
body (linear and angular velocity or acceleration, etc.).
Now these three factors are involved in the principles of
impulse and momentum. And, as noted in Art. 138, in the
analysis of the motion of any mass-system in a plane, three
equations are needed. Two of these equations are obtained by
expressing the principle of linear impulse and linear momentum
with reference to any two rectangular axes in the plane of motion,
and the third equation is obtained by expressing the principle
of angular impulse and angular momentum with reference to
an axis perpendicular to the plane of motion. Thus, in terms
IMPULSE AND MOMENTUM 463
of the symbols already defined, the three equations may be written
as follows :
Qy=AUy,
Lo = AHo.
The particular forms of the expressions for the above quantities
depends upon the type of forces (whether constant or variable,
etc.), the kind of body (whether rigid, etc.), and the type of
motion (whether translation, rotation, plane motion, etc.). The
particular equations for a rigid body having a motion of transla-
tion, of rotation, and of plane motion, under the action of con-
stant forces, are given in the next article.
214. Application of the Principles of Impulse and Momentum
to Special Types of Motion of Rigid Bodies. — The relation between
the impulse of an unbalanced force system which acts on a body
and the momentum of the body, as expressed in a general form by
the three equations in the preceding article, may be expressed
in a more detailed form for the special motions of translation,
rotation, and plane motion of a rigid body as follows :
/. Translation of a Rigid Body under the Action of Constant
Forces. — The velocities of all particles of the body are the same
(Art. 132). Therefore, the linear momentum of the body is
equal to Mv and since its position line passes through the mass-
center of the body, the angular momentum of the body about an
axis through the mass-center is equal to zero. Therefore (using
symbols which have been defined in the preceding articles), the
three equations of Art. 213 become,
llF,-At = A(Mv), = M{v,''-v,'),
^Fy . A^ = A(Mv)y = MiVy" " v/) ,
srAf=o.
77. Rotation of a Rigid Body under the Action of Constant
Forces. — The linear momentum of the body is Mv (Art. 207)
and the angular momentum about the axis of rotation is
7oco (Art. 208). Therefore (using symbols which have been
464 IMPULSE AND MOMENTUM
I defined in the preceding articles) the equations of Art. 213
become,
STo • Af = A(/o(o) = /o(co2 — coi) .
III. Plane Motion of a Rigid Body under the Action of Constant
Forces. — The hnear momentum of the body is Mv (Art. 207) and
the angular momentum of the body about an axis through the
mass-center is /w (Art. 209). Therefore (making use of symbols
which have been defined in the preceding articles), the equations
of Art. 213 become,
l.Fy Lt= L{Mv)y=M{Vy" -Vy'),
SrA^=A(7<o)=7(co2-coi).
It should be noted that each of the three sets of equations above
may be readily transformed into the set of equations which was
derived in Chapter IX for the corresponding motion. However,
under certain conditions, the above equations are more conveni-
ent to use in the analysis and solution of problems than are the
equations of Chapter IX. Furthermore, these equations lead
to an important special principle called the principle of the con-
servation of momentum (Art. 215).
ILLUSTRATIVE PROBLEMS
489. A cylindrical jet of water 1^ in. in diameter impinges on a fixed blade
which is inclined at an angle of 30° with the direction of the jet as shown in
Fig. 444, The velocity of the jet
is 25 ft. per sec. Find the hori-
zontal and vertical components of
the pressure of the water on the
blade (or blade on the water).
~^ V"*'^"^ ^* Assume that the magnitude of the
Fixed blade/ ^^ velocity of the jet is not changed
by the action of the blade. Also
Fig. 444. assume that the only force acting
on the water while it is in contact
with the blade is the pressure of the blade.
Solution. — Let Px and Py be the unknown horizontal and vertical pressures
IMPULSE AND MOMENTUM 465
exerted by the blade on the water, these pressures being the cause of the
change in the momentum of the water.
The principle of impulse and momentum states that,
XFx-At = M{vx"-Vx'), . ....... (1)
l^FyAt=M(Vy"-Vy') (2)
Let At be taken as any convenient time interval (1 sec. say). Then M is the
mass of the water upon which the blade acts in the same time interval.
Taking the direction of the velocity of the impinging water as positive
and the weight of water as 62.5 lb. per cubic foot, we have,
From (1)
Whence,
From (2),
Whence,
,r(1.5)^X25X62.5 ' .
^'^- 4X144X32.2 ^^^ ^5 cos 30 ).
P:c = 0.594(25 -21.65)
= 1.99 lb.
Pj,- 1=0.594(25 sin 30°-0).
P2, = 7.43 1b.
490. A flywheel weighing 1288 lb. is keyed to a shaft 4 in. in diameter.
The shaft transmits a turning moment of 1200 in .-lb. to the flywheel, thereby
increasing its angular velocity from 600 rad. per min. to 50 rad. per sec. in
10 sec. Find the radius of gyration of the flywheel.
Solution. — The flywheel is a rotating rigid body. Only one of the three
equations which apply to a rotating rigid body is needed in this particular
problem; namely,
STo-A« = 7o((02-o)i) ^
= Affco^((02 — tOl).
Hence,
1200^.^ 1288, J_ 600\
-12-X^^ = 32:2^« 1^^-60 j-
Whence,
/co2 = 0.625,
and,
fco = 0.79ft.
PROBLEMS
491. A jet of water 2 in. in diameter has a velocity of 30 ft. per sec. in a
horizontal direction. If the jet impinges normally against a fixed vertical
plane, what is the pressure of the water on the plane? If the plane is moving
in the direction of the jet with a velocity of 10 ft. per sec, what is the pressure
on the plane? Ans. P = 38.1 lb.; P = 16.9 lb.
492. A 55-oz. baseball moving horizontally with a velocity of 150 ft. per
sec. is struck by a bat and is deflected 135° from its original direction as
466
IMPULSE AND MOMENTUM
indicated in Fig. 445. If the speed of the ball as it leaves the bat is 130 ft.
per sec, compute the horizontal and vertical components of the impulse of
the bat on the ball. Assuming that the time
of contact is -^ sec, determine the average
value of the force during the impact.
w 17 = 130 ft./ sec.
,Ball
„cv.. s^ >*r~v\ ^^^' '^^^ table of a planing machine together
yC^v >^ 1^/ ^o. with the material bolted on it weighs 5 tons.
v»y v'=i50 ft./sec. V J J Find the time required to change its velocity
from 20 ft. per min. (cutting stroke) to 40 ft.
per min. in the opposite direction (return stroke)
if the average force of the pinion on the rack
while the velocity is being changed is 200 lb. Ans. i = 1.55 sec.
Fig. 445.
494. The rotating parts of a horizontal-shaft turbine weigh 20 tons
and have a radius of gyration of 2 ft. It takes 10 min. for the turbine to come
to rest from a speed of 55 r.p.m. under the influence of journal friction alone.
The shaft is 12 in. in diameter. What is the average coefficient of friction?
495. A sphere having a weight of 64.4 lb. and a diameter of 30 in. rolls
without slipping down a plane inclined 30° with the horizontal. What will
be the velocity of its center at the end of 5 sec. if the initial velocity of its
center is 30 ft. per sec? Ans. t^ = 87.5 ft. /sec
496. A body slides down a plane inclined 45° with the horizontal. If the
coefficient of kinetic friction is 0.2, how many seconds will it take for the
velocity of the body to change from 10 ft. per sec to 30 ft. per sec?
Ans. f = 1.09 sec.
497. A certain machine gun fires 350 bullets per minute. If each bullet
weighs 1 oz. and the muzzle velocity of the bullets is 2200 ft. per sec, what is
the average reaction of the gun against its support? Neglect the reaction
due to the discharged gases.
498. In the relief valve shown in
Fig. 446 the discharge area is assumed
to be equal to the circumference of the
pipe times the lift of the valve. The
rate of discharge of the water is 2 cu.
ft. per sec The diameter, d, of the
pipe is 6 in. The "lift" is 0.25 in.
The pressure p in the pipe is 30 lb. per
sq. in. Find the force exerted by the
spring on the valve. Hint: The force
causing the change in the horizontal
component of the momentum of the
water from Mvy to Mv^ cos 45° is
the difference between the pressure
on a cross-section of the water in the pipe and the force exerted by the
spring.
Fig. 446.
CONSERVATION OF MOMENTUM 467
215. Conservation of Momentum. — /. Linear Momentum. —
As already noted, the principle of linear impulse and linear
momentum for the motion of any mass-system under the action of
an unbalanced external force system is expressed by the equation.
in which x represents any direction. Now if the resultant of the
forces which act on the body has no component in the a:-direction,
then the impulse, Qx, of the force system in the a:-direction will be
zero and hence, A{Mvx) will be equal to zero. Thus,
A{Mvx) =0 or Mvx = a constant.
That is, if the resultant of the external forces which act on a body
has no component in a given direction, then the linear momentum of
the body in the given direction remains constant. This statement
expresses the principle of the conservation of linear momentum.
II. Angular Momentum. — As already noted, the principle of
angular impulse and angular momentum for the motion of any
body under the action of an unbalanced external force system is
expressed by the equation,
Lo = AHo.
Now if the external forces which act on the body have no resultant
moment about a given axis, 0, then the angular impulse, Lo, of
the forces about the same axis will be zero and hence A^o will be
equal to zero. Thus,
A^o = 0 or Ho = a constant.
That is, if the external forces which act on a body have no re-
sultant moment about an axis, then the angular momentum of the
body with respect to that axis remains constant. This statement
expresses the principle of the conservation of angular momentum.
It was shown in Arts. 208 and 209 that the angular momentum,
Ho, of a body about an axis, 0, is expressed by Iqo) if one of the
following conditions is satisfied: (a) The body is rigid and rotates
about a fixed axis, the 0-axis being taken as the axis of rotation;
(6) the body is rigid and has a plane motion and the 0-axis passes
through the mass-center of the body. Further, Iqco also expresses
468
IMPULSE AND MOMENTUM
the angular momentum of a non-rigid mass-system about a fixed
axis of rotation provided that all parts of the mass-system have
the same angular velocity. Thus, if a rod rotates about a
fixed axis as bodies slide radially outwards (or inwards) along
the rod, the mass-system is not rigid but the angular momentum
of the system about the axis, 0, of rotation is Iqco.
Therefore, the principle of conservation of angular momentum,
when the above conditions are satisfied j may be expressed as follows:
Iqco = a constant.
Thus, if 7o decreases co must increase and vice versa. For example,
a gymnast who leaves the swinging trapeze at the top of a circus
tent with a relatively small angular velocity co (his body being
extended) may increase his angular velocity and make several
complete turns in mid-air as he descends in a vertical plane by
'^ doubling up." His moment of inertia is thereby decreased and
his angular velocity is increased a sufficient amount to keep Iqcc
constant since no external torque acts on him while he is de-
scending.
ILLUSTRATIVE PROBLEMS
499. The weight of the parts of a 3-in. field gun (Fig. 447) which move
during recoil is 950 lb. The weight of the projectile is 15 lb. and that of the
Fig. 447.
powder charge is 1.5 lb. The muzzle velocity is 1700 ft. per sec. Determine
the velocity of free recoil at the time the projectile reaches the muzzle (end of
barrel) assuming that the projectile leaves the gun with a horizontal velocity.
Solution. — Three bodies are to be considered; the projectile, the powder
charge, and the recoiling parts of the gun. Since the recoil is free, there are
no horizontal external forces acting on these three bodies while the projectile
is reaching the muzzle of the gun and hence, the linear momentum of the system
CONSERVATION OF MOMENTUM 469
remains constant. That is, the momentum of the projectile plus the
momentum of the gases is equal to the momentum of the recoiling parts.
Thus,
MpVp +MgVg = MrVr.
The gases (and unburned powder) form a non-rigid body and hence the
velocity, vg, of the mass-center must be used. It is usually assumed that vg is
one-half of the velocity of the projectile. Thus, using weights instead of
masses since they are proportional, we have,
1700
15X1700-1-1.5X^ = 950 vr.
Hence,
25,500+1275
^^ = 950
= 28.1 ft ./sec.
The velocity of free recoil as the projectile reaches the muzzle of the gun is
about 0.7 of the maximum velocity of free recoil. The bore is filled with
gases for a short interval after the projectile leaves the gun and these gases
continue to exert pressure on the breech and thus to increase the velocity of
recoil.
600. Two spheres (Fig. 448) are mounted on a light rod on which the
spheres may slide without friction.
The rod and spheres rotate about the u^ 2- H< ^- H
vertical central axis. A string is , ^ 15' >!< 15' >l
attached to each sphere and runs over [1-y^i[^YtM "^ » 1! ^ {\\ ^r^^ — fl
pulleys so that the pull of each string LHd? '\1J — ^p % "vbAd^^
is directed along the rod. Each sphere
weighs 8 lb. and is 2^ in. in diameter.
When the distance of the center of
each sphere from the axis of rotation
is 2 ft., the angular velocity of the v^^/^;y//y/y^^jj
rod is 60 r.p.m. If the spheres are Fig. 448.
pulled a distance of 6 in. along the rod
toward the axis of rotation what will be the angular velocity of the rod?
Solution. — Since the external forces acting on the spheres have no moment
about the axis of rotation, the angular momentum with respect to the axis of
rotation remains constant. That is, the angular momentum of the spheres
before they are pulled in is equal to their angular momentum after they are
pulled in. Whence,
/l(Ol=/20)2.
Little error will be introduced by considering the spheres to be particles and
by neglecting the mass of the rod. Thus,
2MX(2)2X^5^ = 2MX(1.5)2a)2.
1
470
IMPULSE AND MOMENTUM
Hence,
(02 =
4X27r
2.25
11.15 rad./sec.
= 106.6 r.p.m.
Thus, it will be noted that as the moment of inertia of the spheres decreases,
their angular velocity must increase. And since the moment of inertia
decreases as the square of the distance from the axis of rotation, a relatively
small inward movement of the spheres causes a relatively large increase in the
angular velocity.
If the pulls of the strings are released when the spheres are at any distance,
X (Fig. 449)^ from the axis of rotation,
then no force in the horizontal plane
will be acting on either sphere, and
hence each sphere will continue with
constant speed in a straight line until it
strikes the stop at the end of the rod as
indicated in Fig. 449. That is, the
linear momentum of each sphere remains
constant until the stop is hit and then
changes from mv to mvt. The angular
momentum of the sphere, however, is
the same after the sphere strikes the
stop as it was before the sphere struck
the stop since no moment is introduced
by the pressure of the stop. Thus,
Fig. 449.
mv-x=mvfr.
PROBLEMS
601. If the pulls of the strings on the two spheres of Prob, 500 are released
when each sphere is 2 ft. from the axis of rotation and the angular velocity of
the rod is 90 r.p.m., what will be the linear velocity of each sphere after hitting
the stop, assuming the rod to be 6 ft. long? What is the impulse of the sphere
against the stop?
502. A 2-oz. bullet moving with a velocity of 2000 ft. per sec. strikes,
centrally, a block of wood which is moving on a smooth horizontal plane in the
same direction as the bullet with a velocity of 20 ft. per sec. If the block
of wood in which the bullet embeds itself weighs 16.8 lb., what is the resulting
velocity of the block and bullet? Ans. y = 34.6 ft. /sec.
603. Two similar pulleys are running loose on a shaft. One has an angular
velocity of 10 r.p.m. and the other a velocity of 30 r.p.m. in the opposite
direction. They are suddenly coupled together by means of a friction clutch.
What will be the angular velocity of the pulleys after the clutch has ceased
to slip? What proportion of the kinetic energy is lost?
Ans. 10 r.p.m.; 80 per cent.
IMPACT 471
504. A 4.7 in. howitzer field gun rests on a wooden platform. Recoil is
checked by heavy ropes attached to stakes driven into the ground in front
of the gun. Assuming that the slack in the ropes allows the velocity of free
recoil to be developed, find the velocity of recoil from the following data:
Weight of gun, 7000 lb.; weight of projectile, 63 lb.; weight of powder charge,
6 lb.; muzzle velocity, 1500 ft. per sec.
216. Impact. — The equations of Art. 213 which express the prin-
ciples of impulse and momentum apply to the motion of bodies
whether the bodies move under the action of impulsive forces or of
forces which act during a finite time interval. In fact, as stated
in Art. 201, the principles of impulse and momentum are par-
ticularly well adapted to the solution of kinetics problems which
involve sudden impulses (see Art. 201 for various illustrations).
The effect of impulsive forces on the motion of a body, in most
problems, is so large in comparison with the effect of the other
forces which act on the body that the effect of the other forces on
the motion of the body, while the impact lasts, may be neglected.
The only details of the change in the motion of a body that can be
determined, when the change in the motion is caused by impulsive
forces, are the initial and final velocities of the body. For, the
distance traveled during the impact is indefinitely small; the
time interval is also indefinitely small and hence, the acceleration
produced is indefinitely large since the change in velocity is a
finite quantity. Thus, the distance, time, and acceleration are
indeterminate. There is, however, a definite (appreciable)
change in the velocity although, as just noted, the manner in
which the velocity changes during the period of the impact is
unknown and only the initial and final values of the velocity can
be determined. Therefore, the momentum of the body at the
beginning and at the end of the impact period are definite quantities
and, since these quantities are involved in the principles of impulse
and momentum, problems which involve impulsive forces yield
to this method of solution although the impulse of the impulsive
forces is used and not the forces themselves.
Definitions. — It is convenient to classify or define some of the
ways in which impact of bodies may occur.
Direct and Oblique Impact. — If two bodies colHde and the
velocity of each is directed normal to the striking surfaces, the
impact is said to be direct. When this condition is not fulfilled
the impact is called oblique.
472 IMPULSE AND MOMENTUM
Central and Eccentric Impact. — If two bodies collide in such a
way that the action line of the pressures exerted by the bodies on
each other is directed along the line connecting the mass-centers
of the two bodies, the impact is called central. If the action line
of the pressures is not so directed the impact is called eccentric.
Impact may occur, of course, with two translating bodies;
with two rotating bodies; with two bodies having plane motion;
or, with two bodies, one having any one of these three types of
motion and the other having either one of the remaining two types
of motion. Two of these combinations of motion are treated in
the subsequent articles.
The period of impact may be divided into two parts: (1)
the time of deformation during which the impulsive force is increas-
ing to its maximum value as the two bodies deform, and (2) the
time of restitution during which the bodies are separating and
partially recovering from the deformation. If the two bodies
were perfectly elastic, the period of deformation would be equal
to the period of restitution and, the velocity of separation would be
equal to the velocity of approach. But, all bodies are more or less
inelastic and hence, the velocity of separation is always somewhat
less than the velocity of approach.
Coefficient of Restitution. — For direct central impact of two
bodies, the ratio of the relative velocity of separation to the rela-
tive velocity of approach is defined as the coefficient of restitution.
Thus, if the velocities before impact are denoted by vi and V2
and after impact by Vi' and V2' and if e denotes the coefficient of
restitution, the value of e is defined by the following equation,
g= or V2 —vi = —e{v2 — vi). . . (1)
Experiments show that the value of the coefficient of restitution
for two spheres in central direct impact depends only on the
materials of the two spheres. It is generally assumed that the
value of e as found for two spheres of any two materials is the same
for other bodies of the same materials whether the impact is
central and direct or not. But if the impact is not central and
direct the components of the velocities normal to the impact
surfaces must be used in equation (1) instead of the total veloci-
ties.
Notation. — In dealing with the problem of the impact of^two
IMPACT OF TWO TRANSLATING BODIES 473
bodies as discussed in the following articles, the notation as indi-
cated below will be used.
Ml = the mass of one body.
M2 = the mass of the second body.
vi (or coi) = the linear (or angular) velocity of the impact surface
of Ml before impact occurs.
vo (or 0)2) = the linear (or angular) velocity of the impact surface
of M2 before impact occurs.
Vi' (or coi')=the linear (or angular) velocity of the impact surface
of Ml after impact.
V2 (or C02O = the linear (or angular) velocity of the impact surface
of M2 after impact.
V (or co) = the linear (or angular) velocity of the impact surfaces
at the end of the deformation period.
ta = time of deformation.
^r = time of restitution.
t = td-\-tr = total time of contact.
Pddt = impulse or impact for the time of deformation.
i
I Prdt = impulse or impact for the time of restitution.
Jtd
(Pd) av = the average force for the time of deformation. That is,
X
d
Pddt = {Pdjav ' td'
(Pj.)ao = the average force for the time of restitution. That is,
Prdt={Pr)uv'tr.
f
Jtd
217. Impact of Two Translating «i
Bodies. — Direct Central Impact. — I 1
In Fig. 450 are represented two ^^J^^^.^.JU-y^^^^^^^
translating bodies which collide p^^ ^^q^
with direct central impact. It is
assumed that the values of
Ml, M2, vi, V2, and e,
474 IMPULSE AND MOMENTUM
are known, and it is required to find the values of
V, t^l', f2', I Pddt or {Pd)avtd, and | P4t or {Pr)at'tr.
Evidently five equations must be found from which the five
unknown quantities may be determined. The five equations may
be found as follows: From the principle of conservation of linear
momentum equations (1) and (2) below are obtained. Thus,
MiVi-\-M2V2 = {Mi+M2)v for the period td . . . (1)
{Mi-\-M2)v = MiVi'-\-M2V2' for the period U. . . (2)
And from Art. 216,
vi'-V2'=-e(vi-V2) (3)
From these three equations the values of v, vi, and V2 may be
found. The impulses for the periods td and U may now be found
by applying the principle of linear impulse and momentum for the
periods ta and U, which leads to the following equations,
X
Padt = Mi{v-Vi)=M2{v-V2), (4)
] P4t=Mi{vi'-v)=M2{v2'-v) (5)
PROBLEMS
605. A freight car weighing 40 tons and traveling at a speed of 20 mi. per
hour on a straight track overtakes another car weighing 30 tons and traveling
on the same track in the same direction at a speed of 15 mi. per hour. If
the value of e is 0.2, find the velocity of each car after impact and the impulse
of each car on the other both for the time of deformation and for the time of
restitution.
Ans. yi' = 17.4 mi./hr.; ^2' = 18.4 mi./hr.; I Pdd< = 78201b.-sec.
a*
f
P,.d^ = 15601b.-sec.
606. A body weighing 50 lb. moving to the right collides with a 30-lb.
body moving to the left. The speed of each body is 15 ft. per sec. The
impact of the two bodies is direct and central. If the coefficient of resti-
tution is 0.6, find (a) the velocity of each body after impact, (6) the velocity
IMPACT OF TRANSLATING AND ROTATING BODIES 475
fe
of each body at the end of the deformation period, (c) the impulse or impact
for the deformation period, and (d) the impulse for the period of restitution.
Ans. v/= — 3 ft./sec; y2' = 15 ft./sec; t; = 3.75 ft, /sec.
218. Impact of a Translating Body with a _ "a^-
Rotating Body.— In Fig. 451 let M2 be the
mass of a body which is rotating about a fixed
axis through 0 with angular velocity C02 when
a small translating body of mass Mi having a
velocity vi collides with it at a distance h
from the axis of rotation. This is an example
of direct eccentric impact in which the known
quantities are,
b, h, Ml, M2, I2, vi, C02, V2=^(^2h, and e
J
Wm
Fig. 451.
where I2 denotes the moment of inertia of
M2 about the axis of rotation. And the quantities to be found are,
V, CO, vi', V2'y I ^Pddt, I P4t, I '^Sddt, and | S4t.
Jo Jta Jo Jta
The necessary equations from which to solve for the eight
unknown quantities may be obtained as follows: From the prin-
ciple of conservation of angular momentum the equations (1) and
(2) below are obtained. Thus,
and,
MiVih-^l20i2 = Mivh-\-l2co, for the period ta,
Mivh-{-l2U3 = MiVi'h-\- 120)2', for the period U-
And from Art. 216,
And from Art. 115,
v = o}h. . .
(1)
(2)
(3)
(4)
From these four equations the quantities v, co, vi', and V2 may be
determined. Then from the principle of linear impulse and
momentum for the mass Mi the values of the impulse of the pres-
476 IMPULSE AND MOMENTUM
sure P for the periods ta and U may be found from the following
equations,
P(/d^ = ilfi(z; — z;i) for the period ^d, ... (5)
X
X
Prdt = Mi(vi'-v)iovi]ie\iQv\oiMr. ... (6)
In a similar way by applying the principle of linear impulse and
momentum to the mass M2, the values of the impulse of the reac-
tion S for the periods ta and U may be found from the following
equations,
Hd Hd
I Pddt-\- \ Sddt = M2{v - V2) ior the period td, ... (7)
I Prdt+ \ Srdt = M2{v2 -v)ior the period tr. ... (8)
The kinetics problems involving the impact of two rotating bodies,
of a translating body with a body having plane motion, or of other
combinations of motions may be solved by methods similar to
those used in the last two articles.
PROBLEM
507. The following data apply to the bodies represented in Fig. 451.
rhe weight of the translating body is ^^1=6 lb. and i'i=40 ft. per sec. The
rotating body is a slender rod 6 ft. long which is rotating at 4 rad. per sec.
when impact occurs. The weight of the rod is 15 lb., h = b ft., and e = 0.8.
Xd C^d
Pddt and I Sadt. Assuming that the deformation
period is ru sec, what is the average value of S during the deformation?
Ans. yi' = 20.4ft./sec.; ^2' = 36.4 ft. /sec; /Sap = 5.11b.
219. Loss of Kinetic Energy in Direct Central Impact. — The
linear momentum of two bodies before impact is the same as the
momentum after impact but the kinetic energy of the two bodies
after impact is less than the kinetic energy before impact since a
part of the energy is dissipated during impact in doing work in
compressing the two bodies which are more or less inelastic. If
the value of the coefficient of restitution for two bodies were
unity, that is if the two bodies were perfectly elastic, no energy
would be dissipated during impact since the work done in the
ANGULAR MOMENTUM OF VECTOR QUANTITY 477
deformation period would equal the work recovered during the
period of restitution. The amount of energy lost in impact is
the difference between the energy of the two bodies before impact
and the energy after impact. That is, the loss of kinetic energy,
El, is given by the expression,
Er.={Wivi^+W2V2^)-(iMm'^-^iM2V2'^). . . . (1)
The expression for El, however, is more convenient for use if the
velocities (v'l and v'2) of the two bodies after impact are eliminated.
This may be done as follows: From the principle of conservation
of linear momentum the following equation is found,
MiVi-\-M2V2 = MiVi'-\-M2V2'. .... (2)
And from definition,
vi' — V2'=-e{vi-V2) (3)
By solving equations (2) and (3), the values of vi' and V2' are found
to be,
and.
By substituting these values of vi and V2' in equation (1) the fol-
lowing expression for El results,
It will be noted that for two perfectly elastic bodies (e=l) the
loss of energy in impact is zero and for two plastic bodies (e = 0)
the loss of energy is i^z, = 2 ^ , ^ \v\-v2y.
The loss of energy in direct eccentric impact and in other cases
of impact may be found by a similar method.
220. Angular Momentum a Vector Quantity. — In the preced-
ing articles of this chapter the magnitude, only, of the angular
momentum of a body has been assumed to change. However, in
expressing the principle of angular impulse and angular momentum
for a body which moves so that its plane of motion changes in
direction (such as the propeller of an aeroplane when making a
478
IMPULSE AND MOMENTUM
turn), the angular momentum of the body must be considered as a
quantity having direction as well as magnitude, that is, it must be
considered to be a vector quantity. The angular momentum of a
body may be represented by a vector drawn (1) perpendicular to
the plane of motion of the body to indicate the direction of the
plane of motion, (2) of such a length that it represents, to some
scale, the magnitude of the angular momentum, and (3) with the
sense of rotation indicated by an arrow which points in the direc-
tion along the vector in which a right-handed screw would advance
if given the same sense of rotation as that of the body. Thus,
if a disc (Fig. 452) rotates with angular velocity wi about the axis
OZ, its angular momentum (/^i =7coi) about the axis of rotation is
represented completely by the vec-
tor OB. It will be noted that the
vector representing the angular
momentum may change in length
only, in direction only, or both in
length and in direction. Thus, in
Fig. 452, if the disc is rotated about
the axis OY and at the same time
the angular velocity about the OZ
axis is increased to 0)2, the vector
representing the angular momentum
{H2 = Io)2) of the disc is OC.
Furthermore, the change in the
angular momentum of a body is represented completely by the
change in the angular momentum vector. Thus, in Fig. 452 the
change in the angular momentum of the disc is represented by
the vector BC.
221. The Gyroscopic Couple. — A disc which rolls round a
curved track (Fig. 453a) revolves simultaneously about two rect-
angular axes and hence it has gyroscopic motion (Art. 165).
Gyroscopic motion of a body will here be analyzed briefly by con-
sidering the changes in the angular momentum of the body.
(See Arts. 166 and 167 for another method of analysis.)
In Fig. 453(a) is represented a disc or wheel which rotates with
a constant angular velocity co about its axis (axle) OD as it moves
round the curved track with a constant angular velocity 12. It
will be observed that the angular velocity of the disc about an axis
through D perpendicular to the paper is also equal to fi. At a
Fig. 452.
THE GYROSCOPIC COUPLE
479
given instant, the disc is in the position AiBi and its angular
momentum, Hi, is equal to 7co. After an interval of time dt the
disc is in the position A2B2 and its angular momentum is H2, the
magnitude of which is also equal to Zco. That is, the angular
momentum of the disc (7co) has changed in direction, only, during
the time interval dt.
The change in the angular momentum of the disc from Hi to H2
is represented by the vector EF (Fig. 4536) which connects the
ends of the vectors Hi and H2. Now, since the angle dd is small
Fig. 453.
(greatly exaggerated in Fig. 453) the length of DE, that is, the
magnitude of the change in the angular momentum, is
DE=Hidd = H2dd
= I cade,
and the limiting direction of the vector EF as dS becomes indef-
initely small is perpendicular to Hi. The rate of change of the
angular momentum, then, is
lo^de
-dT^^"^^
and the direction of the vector which represents this rate of change
of the angular momentum is also perpendicular to Hi. Now a
480 IMPULSE AND MOMENTUM
torque or couple is always required to produce a change in
the angular momentum of a body; the moment of the couple is
equal to the rate of change of the angular momentum of the body
(Art. 212) ; the plane in which it acts is perpendicular to the vector
which represents the rate of change of the angular momentum;
and the sense of rotation of the couple is such that it would cause
a right-handed screw to progress (in the direction of the arrow)
along the vector which represents the rate of change of the angular
momentum.
Therefore, a couple C must act on the disc in a plane perpen-
dicular to the plane of the disc and to the plane of the paper, with
a clockwise sense of rotation (as viewed from behind), the mag-
nitude of the couple being
It is evident, therefore, that the disc would turn over counter-
clockwise (outward) unless a couple having a moment equal to
loiQ. acted to prevent the turning. This couple is called the gyro-
scopic couple.
As noted in Art. 165, the forces of the gyroscopic couple fre-
quently cause considerable pressure on the axle of the rotating
body as, for example, in the case of the rotor of an electric loco-
motive when rounding a curve or of the propeller of an aeroplane
when making a turn, etc.
If a body (Fig. 454) rotates or spins about the 2-axis and a
couple Wl having a moment about the x-axis is applied to the body,
the body will rotate about the ^/-axis with an angular velocity 12
unless a couple having a moment about the ^/-axis acts on the body
to prevent the rotation about the 2/-axis. The angular velocitj^ ft
is called the velocity of precession. It is necessary that the body
shall precess in order to develop a resistance to the couple Wl and
hence prevent the disc from falling. This fact may be shown by
following the changes which occur in the angular momentum
about the ^-axis. Thus, when the couple Wl (Fig. 454) first acts,
the z -axis is turned through an angle dd thereby causing a change
AB in the angular momentum of the disc. This change requires
a couple in the horizontal plane with a clockwise sense as viewed
from below. But, since there are no bodies to develop or supply
this couple, the disc turns (precesses) in the horizontal plane and
thus the necessary couple is developed from the inertia of the
THE GYROSCOPIC COUPLE
481
disc. As soon as precession starts, however, that is, as soon as the
s-axis has turned through the angle d(f), the change AC in the
angular momentum is produced. This change requires a couple
in the vertical plane (clockwise as viewed by the reader) to pre-
vent the disc and axle from rotating counter-clockwise in the
vertical plane. This couple is supplied by the two forces W having
a moment Wl. Hence, if the body is allowed to precess about the
Fig. 454.
y-axis with angular velocity fi, the external couple Wl is resisted
by the gyroscopic couple, that is,
W
g
from which the velocity of precession is found to be,
in which k is the radius of gyration of the disc with respect to the
2-axis.
If precession is prevented, a couple (C = 7a;12) must be set up
in the horizontal plane, fi here being the angular velocity, at any
instant, produced by Wl, which will be the same whether the disc
rotates about the 2;-axis or not, since no resistance is offered to the
482 IMPULSE AND MOMENTUM
external couple unless precession is allowed. This explains why a
heavy rotating flywheel or armature on board a ship, with its axis
horizontal and athwartship, will offer no more resistance to the
rolling of the ship than when it is not rotating. The bearing of
the axles, however, must exert a large couple C = IcoQ in a hori-
zontal plane which tends to " nose " the ship around. 12 here
represents the angular velocity of roll.
If an external couple is applied in a horizontal plane to increase
or hurry the precession, the disc and axle (Fig. 454a) will rise
since Iw^>Wl. This principle is employed in the Brennan
mono-rail car, the precession being hurried by the rolling of the
axle of the revolving flywheels, extended, on a shelf attached to
the side of the car. This principle is also used in the " active
type " of gyroscope for stabilizing ships. In this case the pre-
cession is hurried by means of a precession engine which acts after
the ship has rolled a very small amount thus producing a gyro-
scopic righting-couple sufficient to extinguish the roll. Since the
roll is checked in its incipiency, only a small amount of work is
done. The stresses produced in the hull of the ship are also small
for the same reason, and hence, the weight and displacement of
the active type of gyroscope hkewise may be smaU. For those
interested in the various ways in which gyroscopic motion may
arise and in the progress which has been made in the application
of the g3rroscope to the stabihzing of ships, mono-rail cars, and
aeroplanes; to the guiding of torpedoes; to compasses; and to
the Griffin grinding mill, the following references are appended:
1. Journal of the Franklin Institute, May, 1913.
2. American Machinist, August 7 and 14, 1913.
3. Iron Age, December 1, 8, 15, and 22, 1910.
4. Scientific American Supplement, January 26, 1907.
5. Popular Science Monthly, July, 1909.
6. Scientific American, December 18, 1915.
7. Journal American Society of Mechanical Engineers, Novem-
ber, 1920.
Note. — For problems on the gyroscope see Art. 167.
INDEX
The numbers refer to pages
Acceleration
— angular, 242
— components, 236, 245, 269
— linear, 235
— of gravity, 5
— relative, 266
— tangential and normal, 236
— imiform circular, 251
rectilinear, 250
Angle of repose, 133
Angular momentum, 454, 456, 457
— velocity, 229
Anti-resultant, 8
Balancing, 369
— of rotating masses, 370
Bearings, friction in, 145, 148, 443
Bow's notation, 7, 95
Cables, 102, 106
Catenary, 106
Center of gravity, defined, 173
by experiment, 178
parallel force system, 173
oscillation, 364
percussion, 333, 364
suspension, 364
Centrifugal couple, 371
— force, 329, 370
Centroid, defined, 160
— by integration, 163
— graphical method, 177
— of composite figures and bodies,
168
Coefficient of friction, 128
restitution, 472
rolling resistance, 153
speed fluctuation, 435
Composition, defined, 8
— of couples, 60, 62
forces, 28, 29, 34, 48, 53, 56, 65
Compound pendulum, 363, 367
Conservation of energy, 439
momentum, 467
Conservative system, 421, 440
Coriolis' law, 270
Couple, characteristics of, 20
— definition of, 19
— gyroscopic, 382, 478
— moment of, 19
— resolution of, 64
— transformations of, 22, 44
— vector representation of, 20
D'Alembert's principle, 305, 307, 326,
347
Dimensional equations, 25
Displacement
— angular, 224
— linear, 222
— relative, 263
Djoiamics, 2
Djmamometer, 446
Dyne, 297
Efi"ective force, 303
EflSciency, 442
Energy, 420
— conservation of, 439
— dissipation of, 443
— kinetic, 422
— loss of, in impact, 476
— non-mechanical, 428
— of a body, 424
particle, 422
— potential, 421
483
484
INDEX
Equations of motion, defined, 288
— for a particle, 298
plane motion, 339, 340
rotation, 319
translation, 308
Equilibrant, 8
Equilibrium, 8
— algebraic conditions of, 70
— diagrams, 69
— equations of, 69, 72, 73, 78, 82, 110
113, 114
— graphical conditions of, 70
— of coUinear forces, 72
concurrent forces in a plane, 73
— space, 110
parallel forces in a plane, 78
space, 112
non - concurrent, non - parallel
forces in a plane, 82
non - concurrent, non - parallel
forces in space, 114
three forces, 75
Erg, 409, 417
First moment, 160
Flexible cables, 102, 106
Flywheel, stresses in, 356
Force, 1
— characteristics of, 4
— component of, 8
— conception of, 1, 3
— effective, 303
— external effect of, 4
— inertia, 307, 326, 345
— measure of, 4
— moment of, 14
— polygon, 31, 36, 49
— resolution of, 11, 22
— units of, 4
— vector representation of, 6
Forces, classification of, 7
— composition of, 8
— impulsive, 449
— inertia, 326, 345
— resultant of a system of, 8, 28,
— system of, 7
— triangle of, 9
Friction, defined, 127
— belt, 155
— circle, 145
— coefficient of, 128
— cone of, 129
— journal, 145
— laws of, 134
— limiting, 128
— of inclined planes, 135
screws, 142
wedges, 138
— pivot, 148
— rolling, 153
— work lost due to, 443
Funicular polygon, 37
Geepound, 296
Governors, 386
— centrifugal shaft, 391
— inertia shaft, 392
— loaded, 388
— Porter, 389
— Rite's inertia, 393
Guldinus, theorem of, 171
Gyroscope, 378
— analysis of forces in, 379
— precession, axis of, 383
velocity of, 383, 480
Gyroscopic couple, 382, 478
Harmonic motion, 253
Hoop tension, 356
Horsepower, 417
— -hour, 418
Impact, 450, 471
— loss of energy in, 476
Impulse, 450
— angular, 451, 452
— components of, 451
— linear, 451
— moment of, 451, 452
Inertia, defined, 289
— forces, 307, 326, 345
— moment of, 180, 205
— product of, 194
INDEX
485
Instantaneous center,
— of acceleration, 352
velocity, 284
Inclined plane, friction of, 135
Joule, 409
Journal friction, 145
Kilowatt, 417
Kilowatt-hour, 418
Kinematics, 2, 221
Kinetic energy, 420
— lost in impact, 476
— of a body having plane motion, 426
rotation, 425
translation, 424
particle, 422
Kinetics, 2, 287, 303
Lami's theorem, 74
Loss of energy, in bearings, 434
— in impact, 476
Mass, 289
Mass- center, 175
— motion of, 353
Moment-arm, 14, 19
Moment-center, 14, 87
Moment equation, limitations of, 351
Moment of inertia, 180, 205
about inclined axes, 198
, perpendicular planes, 208
by experiment, 220, 367
, composite areas, 190
— , — bodies, 218
, graphical methods, 202
, integration, 184, 209
parallel axis theorem, 183,
207
polar, 181
principal axes, 199
Momentum, 453
— angular, 454, 456, 457
— a vector quantity, 453, 477
— components of, 454
— conservation of, 467
— linear, 454, 455
— moment of, 454, 456, 457
Motion, curvilinear, 222
— gyroscopic, 378, 478
— Newton's laws of, 293
— non-uniformly accelerated, 256
— of a projectile, 248
mass-center of body, 353, 414
rigid bodies, 276
— rectilinear, 222
— relative, 262
— simple harmonic, 253
— uniformly accelerated curvilinear,
251
— uniformly accelerated rectilinear,
250
Motion graphs
— acceleration-time, 261
— distance-time, 228
— speed-time, 241, 260
Newton's laws of motion, 293
Pappus, theorem of, 171
Parabolic cable, 102
Parallelogram law, 8
Particle, 221
— equations of motion for, 297
Pendulum, compound, 363, 367
— conical, 300, 387
— simple, 360
— torsion, 365, 368
Period of oscillation, 362, 364, 366
Pivots, friction of, 148
— loss of energy in, 443
Plane motion, 278, 336
Polar moment of inertia, 181
Pole of force polygon, 36
Poundal, 297
Power, 417
— special equations for, 418
Principal axis, 199
Principle of impulse and momentum,
459, 461
moments, 16, 38, 49, 57, 65
transmissibility, 8
work and energy, 430
Product of inertia, 194
— parallel axis theorem for, 196
486
INDEX
Projectile, motion of, 248
Prony brake, 446
Radius of gyration, 182, 207
Rays of force polygon, 36
Relative motion, 262
Resolution, 8
— of a force, 11
couples, 64
— directions of, 87
Resultant, 8
Resultants of force systems, 28
collinear forces, 28
concurrent forces in a plane, 29
space, 53
non - concurrent, non - parallel
forces in a plane, 48
space, 65
— — parallel forces in a plane, 34
space, 56
Reversed effective force for a body,
307, 326, 345
Rigid body, 3
Rolling resistance, 153
Rotation, 277, 315
Scalar quantity, 6
Screws, 142
Second moment, 180
Sensitiveness of governors, 389
Slug, 296
Space diagram, 6
Speed, 226
Statically indeterminate force sys-
tems, 69
Statics, 2
Stresses in trusses, 88
Strings of funicular polygon, 37
Tangential-effort diagram, 411
Theorems of Pappus and Guldinus,
171
Translation, 276, 305
Triangle law, 9
Trusses, analysis of, 88
— graphical, 95
— method of joints, 89
sections, 90
Units, 294
— absolute system of, 297
— gravitational system of, 297
— systematic system of, 297
Varignon's Theorem, 16
Vector, 6
— diagram, 6
— quantity, 6
Velocity, angular, 229 -
— components of, 231
— linear, 226
— relative, 264
Work and energy, principles of, 430
Work, 406
— done in punching hole in plate, 412
by a force system, 414
— expressions for, 406, 408
— graphical representation of, 409
— units of, 409
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